BASIC CONCEPTS

Circle : A circle is the locus of points which are equidistant from a fixed point and lies on the same plane.

Fixed point is called centre of a circle and constant distance is called radius of the circle

STANDARD EQUATION OF A CIRCLE

The equation of a circle with the centre at $(\mathrm{h},\mathrm{k})$ and radiaus $\mathrm{r}$ is

$(\mathrm{x}-\mathrm{h}{)}^2+(\mathrm{y}-\mathrm{k}{)}^2={\mathrm{r}}^2$

If centre is at the origin and radius is $r$ then the equation of circle is $x^2+y^2=r^2$

GENERALEQUATION OF A CIRCLE

${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$ where $\mathrm{g},\mathrm{f}$, and $\mathrm{c}$ are constants

centre $(-\mathrm{g},-\mathrm{f})$ and radius is $\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}}$

CONDITIONS FOR A SECOND DEGREE EQUATION TO REPRESENT A CIRCLE

$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ is a second degree equation

(i) coefficient of $x^2=$ coefficient of $y^2.$ ie., $a=b$

(ii) $\begin{array}{ll}\text{ coefficient of }xy=0& \text{ ie., }\mathrm{h}=0\end{array}$

If ${\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}>0$ then the circle represents real circle with centre $(-\mathrm{g},-\mathrm{f})$

If ${\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}=0$ then the circle represents point circle since radius is zero

If ${\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}<0$ then the circle is imaginary circle .

EQUATION OF CIRCLE IN VARIOUS FORMS

1. Equation of circle with centre (h.k) and passes through origin. is ${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{hx}+2\mathrm{ky}=0$

Note that when a circle passes through origin the constant term must be zero

2. If the circle touches $x-ax$ is then its equation is $(x\pm h{)}^2+(y\pm k{)}^2=k^2(or)x^2+y^2\pm 2hx\pm 2ky+h^2=0$. In this case radius is ordinate of centre of a circle. Four circles possible

3. If the circle touches $y$-axis then its equation is $(x\pm h{)}^2+(y\pm k{)}^2=h^2($ or $)x^2+y^2\pm 2hx\pm 2ky+k^2=0$. Here radius of the circle is abscissa of the centre. Four circles possible.

4. If the circle touches both the axes then its equation is $(x\pm r{)}^2+(y\pm r{)}^2=r^2$. Four circles possible ${\mathrm{x}}^2+{\mathrm{y}}^2\pm 2\mathrm{rx}\pm 2\mathrm{ry}+{\mathrm{r}}^2=0$

5. If the circle touches $x$-axis at origin then its equation is $x^2+(y\pm k{)}^2=k^2$

${\mathrm{x}}^2+{\mathrm{y}}^2\pm 2\mathrm{ky}=0$

6. If the circle touches $y$-axis at origin then its equation is $(x\pm h{)}^2+y^2=h^2($ or $)x^2+y^2\pm 2hx=0$

7. If the circle passes through origin and cuts intercepts $a$ and $b$ on the axes, then the equation of circle is $x^2+y^2-ax-by=0$ and centre is $c(a/2,b/2)$ four circles possible.

8. Equation of Circle on a given Diameter

If $({\mathrm{x}}_1,{\mathrm{y}}_1)$ and $({\mathrm{x}}_2,{\mathrm{y}}_2)$ are end points of the diameter then the equation of circle is $(\mathrm{x}-{\mathrm{x}}_1)(\mathrm{x}-{\mathrm{x}}_2)+(\mathrm{y}-{\mathrm{y}}_1)(\mathrm{y}-{\mathrm{y}}_2)=0$

9. Parametric form of Circle

$\mathrm{x}=\mathrm{h}+\mathrm{r}\cos \theta$

$\mathrm{y}=\mathrm{k}+\mathrm{r}\sin \theta$

Where $\theta$ is parameter $(0\le \theta \le 2\pi )$

In particular coordinates of any point on the circle $x^2+y^2=r^2$ is $(r\cos \theta ,r\sin \theta )$ on the circle ${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$ is $(-\mathrm{g}+\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}}(\cos \theta ),-f+\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}}(\sin \theta ))$

10. Intercept made on the axes by a circle

Let the equation of circle is $x^2+y^2+2gx+2fy+c=0$

$\mathrm{AB}=\mathrm{x}-$ intercept $=2\sqrt{{\mathrm{g}}^2-\mathrm{c}}$

$\mathrm{CD}=\mathrm{y}-$ intercept $=2\sqrt{{\mathrm{f}}^2-\mathrm{c}}$

Position of a point with respect to a circle .

let the circle is $x^2+y^2+2gx+2fy+c=0$

Point $\mathrm{P}({\mathrm{x}}_1,{\mathrm{y}}_1)$ lies outside, on or inside the circle accordingly $\mathrm{CP}>,=,<$ radius

(or) ${\mathrm{S}}_1={\mathrm{x}}_1{}^2+{\mathrm{y}}_1{}^2+2{\mathrm{gx}}_1+2{\mathrm{fy}}_1+\mathrm{c}>,=,<0$

Maximum and Minimum Distance of a point from the circle

Let the circle ${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$ and point $\mathrm{P}({\mathrm{x}}_1,{\mathrm{y}}_1)$

The maximum and minimum distance from $\mathrm{P}({\mathrm{x}}_1,{\mathrm{y}}_1)$ to the circle are

$\mathrm{PB}=\mathrm{CB}+\mathrm{CP}$

$=\mathrm{r}+\mathrm{CP}$

$\mathrm{PA}=|\mathrm{CP}-\mathrm{CA}|=|\mathrm{PC}-\mathrm{r}|$

$\mathrm{PB}$ is maximum distance and $\mathrm{PA}$ is minimum distance.

Line and a circle

Let $\mathrm{S}=0$ and $\mathrm{L}=0$ be a circle and a line. If $\mathrm{r}$ is the radius of the circle and $\mathrm{d}$ is the length of perpendicular from the centre on the line then,

(i) $\mathrm{d}>\mathrm{r}\Rightarrow$ line does not meet circle

(ii) $\mathrm{d}=\mathrm{r}\Rightarrow$ line touches the circle. It is a tangent to the circle

(iii) $\mathrm{d}<\mathrm{r}\Rightarrow$ line intersect the circle line is a secant to the circle

(iv) $\mathrm{d}=0\Rightarrow$ line is a diameter of the circle

If $y=mx+c$ is a line and $x^2+y^2=r^2$ is a circle then

(i) ${\mathrm{c}}^2>{\mathrm{r}}^2(1+{\mathrm{m}}^2)\Rightarrow$ The line is a secant of the circle. The line intersects the circle in two distinct points.

(ii) ${\mathrm{c}}^2={\mathrm{r}}^2(1+{\mathrm{m}}^2)\Rightarrow$ The line is a tangent to the circle. The line touches the circle at unique point.

(iii) ${\mathrm{c}}^2<{\mathrm{r}}^2(1+{\mathrm{m}}^2)\Rightarrow$ The line does not meet the circle

Equations of Tangents

Point form :

Equation of the tangent to the circle $x^2+y^2=a^2$ at the point $(x_1,y_1)$ on it is : $x_1+y_1=a^2$ Equation of the tangent to the circle $x^2+y^2+2gx+2fy+c=0$ at the point $(x_1,y_1)$ on it is ${\mathrm{xx}}_1+{\mathrm{yy}}_1+\mathrm{g}(\mathrm{x}+{\mathrm{x}}_1)+f(\mathrm{y}+{\mathrm{y}}_1)+\mathrm{c}=0$

Parametric forms

Equation of the tangent to the circle $x^2+y^2=a^2$ at the point $(a\cos \theta \cdot \mathrm{asin}\theta )$ on it is $x\cos \theta +\mathrm{y}\sin \theta =\mathrm{a}$

Slope form

The equation of a tangent of slope $m$ to the circle $x^2+y^2=a^2$ is $y=mx\pm a\sqrt{1+m^2}$

The coordinates of the point of contact are $(\pm {\displaystyle \frac{\mathrm{am}}{\sqrt{1+{\mathrm{m}}^2}}},\overline{+}{\displaystyle \frac{\mathrm{a}}{\sqrt{1+{\mathrm{m}}^2}}})$

(i) Condition for a line $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ to be a tangent to the circle ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2$ is ${\mathrm{c}}^2={\mathrm{a}}^2(1+{\mathrm{m}}^2)$ or $\mathrm{c}=$ $\pm \mathrm{a}\sqrt{1+{\mathrm{m}}^2}$

(ii) Condition that the line $\ell \mathrm{x}+\mathrm{my}+\mathrm{n}=0$ touches the circle ${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$ is $(\ell \mathrm{g}+\mathrm{mf}+\mathrm{n}{)}^2$ $=({\ell }^2+{\mathrm{m}}^2)({\mathrm{g}}^2+f^2-\mathrm{c})$

(iii) Equation of tangent to the circle $x^2+y^2+2gx+2fy+c=0$ in terms of slope is $y=mx+mg-$ ${\mathrm{f}}_{\pm }\sqrt{{\mathrm{g}}^2+f^2-\mathrm{c}}\sqrt{1+{\mathrm{m}}^2}$

(iv) The line $\ell \mathrm{x}+\mathrm{my}+\mathrm{n}=0$ touches the circle $(\mathrm{x}-\mathrm{a}{)}^2+(\mathrm{y}-\mathrm{b}{)}^2={\mathrm{r}}^2$ if $(\mathrm{a}\ell +\mathrm{bm}+\mathrm{n}{)}^2={\mathrm{r}}^2({\ell }^2+{\mathrm{m}}^2)$

(v) If the line $y=mx+c$ is the tangent to the circle $x^2+y^2=r^2$ then point of contact is given by $({\displaystyle \frac{-m{r}^2}{c}},{\displaystyle \frac{r^2}{c}})$

(vi) If the line $ax+by+c=0$ is the tangent to the circle $x^2+y^2=r^2$ then point of contact is given by $({\displaystyle \frac{-{\mathrm{ar}}^2}{\mathrm{c}}},{\displaystyle \frac{-{\mathrm{br}}^2}{\mathrm{c}}})$

Tangents from a point outside the circle

If circle is ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2$ and any tangent to the circle is

$\mathrm{y}=\mathrm{mx}+\mathrm{a}\sqrt{1+{\mathrm{m}}^2}$

$\mathrm{P}({\mathrm{x}}_1,{\mathrm{y}}_1)$ lies on the tangent

$\therefore {\mathrm{y}}_1={\mathrm{mx}}_1+\mathrm{a}\sqrt{1+{\mathrm{m}}^2}$

${\mathrm{y}}_1-{\mathrm{mx}}_1=\mathrm{a}\sqrt{1+{\mathrm{m}}^2}$

Squaring

${({\mathrm{y}}_1-{\mathrm{mx}}_1)}^2={\mathrm{a}}^2(1+{\mathrm{m}}^2)$

${\mathrm{m}}^2({\mathrm{x}}_1{}^2-{\mathrm{a}}^2)-2\mathrm{mx}{\mathrm{y}}_1+{\mathrm{y}}_1{}^2-{\mathrm{a}}^2=0$ is a quadratic equation in $\mathrm{m}$ which gives 2 values of $\mathrm{m}$.

We get two equations of tangents.

Let the circle be ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{r}}^2$ and a point $\mathrm{P}\left({\mathrm{x}}_{11}{\mathrm{y}}_1\right)$ outside the circle .

Let the slope of the tangent is $m$ then equation of the tangent is $\mathrm{y}-{\mathrm{y}}_1=\mathrm{m}(\mathrm{x}-{\mathrm{x}}_1)$

Now find the distance of this line from the centre $(0,0)$ and equate to the radius. We get equation in $\left|{\displaystyle \frac{{\mathrm{mx}}_1-{\mathrm{y}}_1}{\sqrt{1+{\mathrm{m}}^2}}}\right|=\mathrm{r}$

Length of the tangent from a point to a circle.

Let the circle be ${\mathrm{S}}_{\equiv {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2f\mathrm{y}+\mathrm{c}}=0$ then centre and radius of circle are $(-\mathrm{g},-f)$ and

$\sqrt{{\mathrm{g}}^2+f^2-\mathrm{c}}$ respectively and let $\mathrm{P}({\mathrm{x}}_1,{\mathrm{y}}_1)$ be any point outside the circle.

$\mathrm{PT}=\sqrt{x_1^2+y_1^2+2g{x}_1+2f{y}_1+c}$

$=\sqrt{{\mathrm{S}}_1}$

Power of point with respect to a circle

The power of $\mathrm{P}({\mathrm{x}}_1,{\mathrm{y}}_1)$ with respect to $\mathrm{S}\equiv$

$x^2+y^2+2gx+2fy+c=0$ is equal to PA. PB or PC.PD which is ${\mathrm{x}}_1{}^2+{\mathrm{y}}_1{}^2+2{\mathrm{gx}}_1+2f{\mathrm{y}}_1+\mathrm{c}=0\Rightarrow$ ${\mathrm{S}}_1=0(:\mathrm{PA}.\mathrm{PB}=\mathrm{PC}.\mathrm{PD}={\mathrm{PT}}^2)$

Note :

(i) The power of the point outside the circle is positive

(ii) The power of the point on the circle is zero

(iii) The power of the point inside the circle is negative

Pair of tangents

The equation of the pair of tangents drawn from the point $\mathrm{P}({\mathrm{x}}_1,{\mathrm{y}}_1)$ to the circle $\mathrm{S}=0$ is ${\mathrm{SS}}_1={\mathrm{T}}^2$

Where $S=x^2+y^2+2gx+2fy+c$

$\begin{array}{rl}& {\mathrm{S}}_1={\mathrm{x}}_1^2+{\mathrm{y}}_1^2+2{\mathrm{gx}}_1+2f{\mathrm{y}}_1+\mathrm{c}\\ & \mathrm{T}={\mathrm{xx}}_1+{\mathrm{yy}}_1+\mathrm{g}(\mathrm{x}+{\mathrm{x}}_1)+f({\mathrm{y}}_1+{\mathrm{y}}_1)+\mathrm{c}\end{array}$

Note : The pair of tangents from $(0,0)$ to the circle $x^2+y^2+2gx+2fy+c=0$ are at right angles if ${\mathrm{g}}^2+{\mathrm{f}}^2=2\mathrm{c}$

Normal to a circle at a given point

The normal of a circle at any point is a straight line which is perpendicular to the tangent at the point and always passes through the centre of the circle.

Point form

To find the equation of normal to the circle $x^2+y^2$

$={\mathrm{a}}^2$ at the point $\mathrm{p}({\mathrm{x}}_1,{\mathrm{y}}_1)$ on it .

Since we know that normal passes through the centre of a circle. So we get two points on normal using two point form of a line we get the equation of normal as

${\displaystyle \frac{\mathrm{y}-0}{{\mathrm{y}}_1-0}}={\displaystyle \frac{\mathrm{x}-0}{{\mathrm{x}}_1-0}}$

or

${\displaystyle \frac{\mathrm{y}}{{\mathrm{y}}_1}}={\displaystyle \frac{\mathrm{x}}{{\mathrm{x}}_1}}$

${\mathrm{xy}}_1-{\mathrm{yx}}_1=0$

To find normal at $({\mathrm{x}}_1,{\mathrm{y}}_1)$ of scored degree conics ${\mathrm{ax}}^2+2\mathrm{hxy}+\mathrm{by}{\mathrm{y}}^2+2\mathrm{gx}+2f\mathrm{y}+\mathrm{c}=0$ _________________(1)

then according to determinant $\left|\begin{array}{lll}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|$

Write first two rows as ${\mathrm{ax}}_1+\mathrm{h}{\mathrm{y}}_1+\mathrm{g}\text{ and }\mathrm{h}{\mathrm{x}}_1+\mathrm{b}{\mathrm{y}}_1+\mathrm{f}$

Then normal at $({\mathrm{x}}_1,{\mathrm{y}}_1)$ of (1) is ${\displaystyle \frac{\mathrm{x}-{\mathrm{x}}_1}{{\mathrm{ax}}_1+{\mathrm{hy}}_1+\mathrm{g}}}={\displaystyle \frac{\mathrm{y}-{\mathrm{y}}_1}{{\mathrm{hx}}_1+{\mathrm{by}}_1+\mathrm{f}}}$

If equation of circle is $x^2+y^2=a^2$

here $\mathrm{a}=\mathrm{b}=1$ and $\mathrm{h}=0=\mathrm{g}=\mathrm{f}$

$\therefore {\displaystyle \frac{\mathrm{x}-{\mathrm{x}}_1}{{\mathrm{x}}_1}}={\displaystyle \frac{\mathrm{y}-{\mathrm{y}}_1}{{\mathrm{y}}_1}}$

${\displaystyle \frac{\mathrm{x}}{{\mathrm{x}}_1}}-1={\displaystyle \frac{\mathrm{y}}{{\mathrm{y}}_1}}-1$

${\displaystyle \frac{\mathrm{x}}{{\mathrm{x}}_1}}={\displaystyle \frac{\mathrm{y}}{{\mathrm{y}}_1}}$ is equation of normal at $({\mathrm{x}}_1,{\mathrm{y}}_1)$

If equation of circle is $x^2+y^2+2gx+2fy+c=0$

here $\mathrm{a}=\mathrm{b}=1$ and $\mathrm{h}=0$

Then ${\displaystyle \frac{x-x_1}{x_1+g}}={\displaystyle \frac{y-y_1}{y_1+f}}$

Chord of contact

Let the equation of circle be $x^2+y^2=r^{2}PA$ and $PB$ are pair of tangents drawn from the point $P(x_1,y_1)$ then $\mathrm{AB}$ is the chord of contact with $\mathrm{A}$ and $\mathrm{B}$ as its points of contact.

$\therefore$ Equation of chord of contact $\mathrm{AB}$ is ${\mathrm{xx}}_1+{\mathrm{yy}}_1={\mathrm{a}}^2$ Equation of chord of contact look like equation of tangent at point but point are different If the equation of circle be $x^2+y^2+2gx+2fy+c=$ 0 then the equation of chord of contact is ${\mathrm{xx}}_1+{\mathrm{yy}}_1+\mathrm{g}(\mathrm{x}+{\mathrm{x}}_1)+\mathrm{f}(\mathrm{y}+{\mathrm{y}}_1)+\mathrm{c}=0$

Equation of the chord bisected at a given point

Let the equation of circle be ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{r}}^2$ and $\mathrm{AB}$ is a chord of it $\mathrm{Let}\mathrm{M}({\mathrm{x}}_1,{\mathrm{y}}_1)$ be midpoint of $\mathrm{AB}$.

$\begin{array}{rl}& \text{ Slope of }\mathrm{CM}={\displaystyle \frac{{\mathrm{y}}_1}{{\mathrm{x}}_1}}\\ & \text{ Slope of }\mathrm{AB}={\displaystyle \frac{-{\mathrm{x}}_1}{{\mathrm{y}}_1}}\end{array}$

Equation of chord $\mathrm{AB}$ is

$\begin{array}{rl}& \mathrm{y}-{\mathrm{y}}_1={\displaystyle \frac{-{\mathrm{x}}_1}{{\mathrm{y}}_1}}(\mathrm{x}-{\mathrm{x}}_1)\\ & {\mathrm{yy}}_1-{\mathrm{y}}_1^2=-{\mathrm{xx}}_1+{\mathrm{x}}_1^2\\ & {\mathrm{xx}}_1+{\mathrm{yy}}_1={\mathrm{x}}_1^2+{\mathrm{y}}_1^2\\ & {\mathrm{xx}}_1+{\mathrm{yy}}_1-{\mathrm{a}}^2=({\mathrm{x}}_1^2+{\mathrm{y}}_1^2-{\mathrm{a}}^2)\\ & \mathrm{T}={\mathrm{S}}_1\end{array}$

If the equation of circle be $x^2+y^2+2gx+2fy+c=0$ then the equation of chord which is bisected at $(x_1$, ${\mathrm{y}}_1$ ) is

${\mathrm{xx}}_1+{\mathrm{yy}}_1+\mathrm{g}(\mathrm{x}+{\mathrm{x}}_1)+\mathrm{f}(\mathrm{y}+{\mathrm{y}}_1)+\mathrm{c}={\mathrm{x}}_1^2+{\mathrm{y}}_1^2+2{\mathrm{gx}}_1+{\mathrm{fy}}_1+\mathrm{c}$

Director circle and its equation

The locus of the point of intersection of two perpendicular tangents to a given circle is known as its director circle.

Equation of Director Circle

Let $\mathrm{P}(\mathrm{h},\mathrm{k})$ be the point of intersection of tangents to a circle ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{r}}^2$ at right angle.

$\mathrm{ACBP}$ is a square

$\therefore \mathrm{AC}=\mathrm{CP}\sin {45}^{\circ }$

$\mathrm{r}={\displaystyle \frac{\sqrt{{\mathrm{h}}^2+{\mathrm{k}}^2}}{\sqrt{2}}}$

Squaring we get

$2{\mathrm{r}}^2={\mathrm{h}}^2+{\mathrm{k}}^2$

or ${\mathrm{x}}^2+{\mathrm{y}}^2=2{\mathrm{r}}^2$ is the required equation. of director circle

Intersection of two circles, common Tangents to two circles

Let the two circles be ${(\mathrm{x}-{\mathrm{g}}_1)}^2+{(\mathrm{y}-{\mathrm{f}}_1)}^2={\mathrm{r}}_1{}^2$ and ${(\mathrm{x}-{\mathrm{g}}_2)}^2+{(\mathrm{y}-{\mathrm{f}}_2)}^2={\mathrm{r}}_2{}^2$

with centres ${\mathrm{C}}_1({\mathrm{g}}_1,{\mathrm{f}}_1)$ and ${\mathrm{C}}_2({\mathrm{g}}_2,{\mathrm{f}}_2)$ and radii ${\mathrm{r}}_1$ and ${\mathrm{r}}_2$ respectively

Different cases of intersection of two circles

Case I When $\left|{\mathrm{C}}_1{\mathrm{C}}_2\right|>{\mathrm{r}}_1+{\mathrm{r}}_2$

ie., distance between the centre is greater than sum of the radii.

In this case there are 4 common tangents can be drawn.

Two direct common tangents (circles lies on the same side of the tangent)

$\mathrm{ABE}$ and $\mathrm{CDE}$

Two indirect (Transverse) common tangents (circles lies opposite side of the tangent) GFH & IFJ.

Note that centres of two circles and point of intersection of tangents are collinear also

${\displaystyle \frac{{\mathrm{C}}_1\mathrm{E}}{{\mathrm{C}}_2\mathrm{E}}}={\displaystyle \frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}}$ & ${\displaystyle \frac{{\mathrm{C}}_1\mathrm{F}}{{\mathrm{C}}_2\mathrm{F}}}={\displaystyle \frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}}$

To find the equations of common tangents.

Let us assume equation of tangent of any circle in slope form be

$(\mathrm{y}+\mathrm{f})=\mathrm{m}(\mathrm{x}+\mathrm{g})+\mathrm{r}\sqrt{1+{\mathrm{m}}^2}$

Points $\mathrm{E}$ & $\mathrm{F}$ satisfy this equation. Substitute the coordinates of $\mathrm{E}$ & $\mathrm{F}$ to get the values of $\mathrm{m}$.

Substitnde the values of $m$ we get required respective common tangent equation

Case II When $\left|{\mathrm{C}}_1{\mathrm{C}}_2\right|={\mathrm{r}}_1+{\mathrm{r}}_2$

ie., distance between the centre is equal to sum of the radii.

In this case there are 3 common tangents 2 direct common tangent and one transverse common tangent

The equation of tangent at point $\mathrm{F}$ is ${\mathrm{S}}_1-{\mathrm{S}}_2=0$ where ${\mathrm{S}}_1=0$ and ${\mathrm{S}}_2=0$ are equations of circles.

Coordinate of $F$ are $({\displaystyle \frac{r_1{g}_2+r_2{g}_1}{r_1+r_2}},{\displaystyle \frac{r_1{f}_2+r_2{f}_1}{r_1+r_2}})$

Case III When $\left|{\mathrm{C}}_1{\mathrm{C}}_2\right|<{\mathrm{r}}_1+{\mathrm{r}}_2$

ie,. Distance between the centre is less than sum of the radii.

In this case only two direct common tangents are possible.

Case IV When $\left|{\mathrm{C}}_1{\mathrm{C}}_2\right|=|{\mathrm{r}}_1-{\mathrm{r}}_2|$

ie., distance between the centre is equal to difference of the radii.

Then the two circles touch each other internally.

In this case only one direct common tangent is possible.

Equation of common tangent is ${\mathrm{S}}_1-{\mathrm{S}}_2=0$

$F$ divides line joining ${\mathrm{C}}_1$ and ${\mathrm{C}}_2$ externally in the ratio ${\mathrm{r}}_1:{\mathrm{r}}_2$

$\therefore$ coordinates of $F$ are $({\displaystyle \frac{{\mathrm{r}}_1{\mathrm{g}}_2-{\mathrm{r}}_2{\mathrm{g}}_1}{{\mathrm{r}}_1-{\mathrm{r}}_2}},{\displaystyle \frac{{\mathrm{r}}_1{\mathrm{f}}_2-{\mathrm{r}}_2{\mathrm{f}}_1}{{\mathrm{r}}_1-{\mathrm{r}}_2}})$

Case V When $\left|{\mathrm{C}}_1{\mathrm{C}}_2\right|<|{\mathrm{r}}_1-{\mathbf{r}}_2|$

ie., distance between the centre is less than the difference of the radii. Then one circle contains the other

In this case there is no real Common tangents.

Length of direct common tangent if $\left|{\mathbf{C}}_1{\mathbf{C}}_2\right|>{\mathbf{r}}_1+{\mathbf{r}}_2$

length of direct common tangent $=\sqrt{{\mathrm{d}}^2-{({\mathrm{r}}_1-{\mathrm{r}}_2)}^2}$

length of transverse common tangent $=\sqrt{{\mathrm{d}}^2-{({\mathrm{r}}_1+{\mathrm{r}}_2)}^2}$

Where $\mathrm{d}=\left|{\mathrm{C}}_1{\mathrm{C}}_2\right|$ and ${\mathrm{r}}_1,{\mathrm{r}}_2$ are radii of the circles

Common Chord of two circles

The common chord joining the point of intersection of two circles is called their common chord .IfS $=0$ and ${\mathrm{S}}^1=0$ be two intersecting circles then the equation of their common chord is $\mathrm{S}-{\mathrm{S}}^1=0$

Let the equations of circles are

$\mathrm{S}={\mathrm{x}}^2+{\mathrm{y}}^2+2{\mathrm{g}}_1\mathrm{x}+2{\mathrm{f}}_1\mathrm{y}+{\mathrm{c}}_1=0$

${\mathrm{S}}^1={\mathrm{x}}^2+{\mathrm{y}}^2+2{\mathrm{g}}_2\mathrm{x}+2{\mathrm{f}}_2\mathrm{y}+{\mathrm{c}}_2=0$

Then equation of common chord $\mathrm{AB}$ is

$\mathrm{S}-{\mathrm{S}}^1=0$

$2\mathrm{x}({\mathrm{g}}_1-{\mathrm{g}}_2)+2\mathrm{y}({\mathrm{f}}_1-{\mathrm{f}}_2)+{\mathrm{c}}_1-{\mathrm{c}}_2=0$

Length of common chord is $2\mathrm{AM}=\sqrt{{\left({\mathrm{C}}_1\mathrm{A}\right)}^2-{\left({\mathrm{C}}_1\mathrm{M}\right)}^2}$

${\mathrm{C}}_1\mathrm{M}$ is the length of perpendicular from the centre ${\mathrm{C}}_1$ to common chord and ${\mathrm{C}}_1\mathrm{A}$ is radius of circle.

Note :

(i) Common chord $\mathrm{AB}$ becomes maximum length when it is a diameter of the smaller one.

(ii) Circle on the common chord a diameter then centre of the circle passing through A and B lie on the common chord of the two circle

(iii) If the length of common chord is zero then the two circles touch each other and the common chord be comes the common tangent to the two circles at the point of contact.

Angle of intersection of two circles

Let the two circles $S\equiv x^2+y^2+2gx+2fy+c=0$ and $S^{\prime }\equiv x^2+y^2+2g_{1}x+2f_{1}y+c_1=0$ intersect each other at the point $P$ and $Q$. The angle $\theta$ between two circles $S=0$ and $S^{\prime }=0$ is defined as the angle between the tangents to the two circles at the point of intersection. $\theta$ must be taken acute angle .

${\mathrm{C}}_1$ and ${\mathrm{C}}_2$ are the centres of circles $\mathrm{S}=0$ and ${\mathrm{S}}^{\prime }=0$ then ${\mathrm{C}}_1(-\mathrm{g},-\mathrm{f})$ and ${\mathrm{C}}_2(-{\mathrm{g}}_1,-{\mathrm{f}}_1)$ and their radii ${\mathrm{r}}_1=\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}}$ & ${\mathrm{r}}_2=\sqrt{{\mathrm{g}}_1{}^2+{\mathrm{f}}_1{}^2-{\mathrm{c}}_1}$

Let $\mathrm{d}=\left|{\mathrm{c}}_1{\mathrm{c}}_2\right|=\sqrt{{(\mathrm{g}-{\mathrm{g}}_1)}^2+{(\mathrm{f}-{\mathrm{f}}_1)}^2}=\sqrt{{\mathrm{g}}^2+{\mathrm{g}}_1{}^2-2{\mathrm{gg}}_1+{\mathrm{f}}^2+{\mathrm{f}}_1{}^2-2{\mathrm{ff}}_1}$

${\mathrm{C}}_1\mathrm{P}\perp {\mathrm{AA}}^{\prime },{\mathrm{C}}_2\mathrm{P}\perp {\mathrm{BB}}^{\prime }$ since radius is perpendicular to the tangent at the point of contact

ie. $\mathrm{\angle }{\mathrm{C}}_1{\mathrm{PA}}^{\prime }=90$ and $\mathrm{\angle }{\mathrm{C}}_2{\mathrm{PB}}^{\prime }=90$

$\therefore \mathrm{\angle }{\mathrm{C}}_1{\mathrm{PB}}^{\prime }=90-\theta$ and $\mathrm{\angle }{\mathrm{C}}_2{\mathrm{PA}}^{\prime }=90-\theta$

Hence $\mathrm{\angle }{\mathrm{C}}_1{\mathrm{PC}}_2=90-\theta +\theta +90-\theta =180-\theta$

Now in ${}_{\mathrm{\Delta }}{\mathrm{C}}_1{\mathrm{PC}}_2$

$\cos (180-\theta )={\displaystyle \frac{{\mathrm{r}}_1{}^2+{\mathrm{r}}_2{}^2-{\mathrm{d}}^2}{2{\mathrm{r}}_1{\mathrm{r}}_2}}$ (cosine rule)

$\cos \theta =\left|{\displaystyle \frac{{\mathrm{r}}_1{}^2+{\mathrm{r}}_2{}^2-{\mathrm{d}}^2}{2{\mathrm{r}}_1{\mathrm{r}}_2}}\right|$

If the angle between the circles is 90 ie., $\theta =90$, then $\cos 90=0$ Then the circles are said to be orthogonal circles or the circles cut each other orthogonally.

$\therefore {\mathrm{r}}_1^2+{\mathrm{r}}_2^2-{\mathrm{d}}^2=0$

${\mathrm{g}}^2+{\mathrm{f}}^2-{\mathrm{c}}^2+{\mathrm{g}}_1{}^2+{\mathrm{f}}_1^2-{\mathrm{c}}_1-{\mathrm{g}}^2-{\mathrm{g}}_1{}^2+2{\mathrm{gg}}_1-{\mathrm{f}}^2-{\mathrm{f}}_1{}^2+2{\mathrm{ff}}_1=0$

$2{\mathrm{gg}}_1+2{\mathrm{ff}}_1=\mathrm{c}+{\mathrm{c}}_1$

It is a condition for two circles to be orthogonal

Radical axis

The radical axis of two circles is the locus of a point which moves such that the lengths of the tangents drawn from it to the two circles are equal.

Consider $\mathrm{S}={\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$ and ${\mathrm{S}}^{\prime }={\mathrm{x}}^2+{\mathrm{y}}^2+2{\mathrm{g}}_1\mathrm{x}+2{\mathrm{f}}_1\mathrm{y}+{\mathrm{c}}_1=0$

Let $\mathrm{P}({\mathrm{x}}_1,{\mathrm{y}}_1)$ be any point such that $\left|{\mathrm{PT}}_1\right|=\left|{\mathrm{PT}}_2\right|$

or ${\left({\mathrm{PT}}_1\right)}^2={\left({\mathrm{PT}}_2\right)}^2$

${\mathrm{x}}_1{}^2+{\mathrm{y}}_1{}^2+2{\mathrm{gx}}_1+2{\mathrm{fy}}_1+\mathrm{c}={\mathrm{x}}_1{}^2+{\mathrm{y}}_1{}^2+2{\mathrm{g}}_1{\mathrm{x}}_1+2{\mathrm{f}}_1{\mathrm{y}}_1+{\mathrm{c}}_1$

$2{\mathrm{x}}_1(\mathrm{g}-{\mathrm{g}}_1)+2{\mathrm{y}}_1(\mathrm{f}-{\mathrm{f}}_1)+\mathrm{c}-{\mathrm{c}}_1=0$ _____________________(1)

$\therefore$ locus of $\mathrm{P}({\mathrm{x}}_1,{\mathrm{y}}_1)$ is a straight line and equation (1) represents its equation.

Equation of radical axis is $2x(g-g_1)+2y(f-f_1)+c-c_1=0$

Properties of radical axis

1. Radical axis is perpendicular to the line joining the centres of the given circles.

(Slope of radical axis) $\times$ slope of line joining centres $=-1$

${\displaystyle \frac{-(g-g_1)}{(f-f_1)}}\times {\displaystyle \frac{f-f_1}{g-g_1}}=-1$

2. The radical axis bisects common tangents of two circles

As $\mathrm{M}$ lies on radical axis and $\mathrm{AB}$ is a tangent to the circles $\therefore \mathrm{AM}=\mathrm{BM}$.

Hence radical axis bisects the common tangents

3. If two circles cut a third circle orthogonally, then the radical axis of the two circles will pass through the centre of the third circle, or the locus of the centre of a circle cutting two given circles orthogonally is the radical axis of the given two circles

$\mathrm{CP}=\mathrm{CQ}$

Hence $\mathrm{C}$ lies on the radical axis of the circles $\mathrm{S}=0$ and ${\mathrm{S}}^{\prime }=0$

4. The position of the radical axis of the two circles geometrically

Radical Centre

The radical axes of three circles, taken in pairs, meet in a point, which is called their radical centre.

Let the three circles be $S_1=0,S_2=0S_3=0$

OL, OM, ON be radical axes of the pair sets of circles $\{{\mathrm{S}}_1=0,{\mathrm{S}}_2=0\},\{{\mathrm{S}}_2=0,{\mathrm{S}}_3=0\},\{{\mathrm{S}}_3=0$, ${\mathrm{S}}_1=0\}$ respectively

Equations of OL, $\mathrm{OM}$ and $\mathrm{ON}$ are ${\mathrm{S}}_1-{\mathrm{S}}_2=0{\mathrm{S}}_2-{\mathrm{S}}_3=0$ and ${\mathrm{S}}_3-{\mathrm{S}}_1=0$

Family of lines passes through point of intersection of lines $S_1-S_2=0$ and $S_2-S_3=0$ is

${\mathrm{S}}_1-{\mathrm{S}}_2+\lambda ({\mathrm{S}}_2-{\mathrm{S}}_3)=0$

If $\lambda =1$ then the line becomes $S_3-S_1=0$

3. The radical centre of three given circles will be the centre of a fourth circle which cuts all the three circles orthogonally. and the radius of the fourth circle is the length of tangent drawn from radical centre of the three given circles to any of these circles $(\mathrm{x}-\mathrm{y}{)}^2+(\mathrm{y}-\mathrm{k}{)}^2={\mathrm{r}}^2$ is the fourth circle with centre $(\mathrm{h},\mathrm{k})$ and radius $\mathrm{r}$ Centre $(h,k)$ is the radical centre of three circles and radius $r$ is the length of tangent to one of these three circles from radical centre $(\mathrm{h},\mathrm{k})$

Co - Axial System of Circles

A system of circles or family of circles, every pair of which have the same radical axis are called co-axial circles

1. The equation of family of co-axial circles when the equation of radical axis and one circle are given

$\mathrm{L}=\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$

$\mathrm{S}\equiv {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$

Then equation of co-axial circle is $\mathrm{S}+\lambda \mathrm{L}=0$

2. The equation of co-axial system of a circles where the equation of any two circles of the system are

${\mathrm{S}}_1={\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$

${\mathrm{S}}_2={\mathrm{x}}^2+{\mathrm{y}}^2+2{\mathrm{g}}_1\mathrm{x}+2{\mathrm{f}}_1\mathrm{y}+{\mathrm{c}}_1=0$

respectively is ${\mathrm{S}}_1+\lambda ({\mathrm{S}}_1-{\mathrm{S}}_2)=0$

and ${\mathrm{S}}_2+\lambda ({\mathrm{S}}_1-{\mathrm{S}}_2)=0(\lambda \ne -1)$

${\mathrm{S}}_1+\lambda {\mathrm{S}}_2=0(\lambda \ne -1)$

${\mathrm{S}}_1+\lambda {\mathrm{S}}_2=0(\lambda \ne -1)$

3. The equation of a system of co-axial circles in the simplest form is ${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+\mathrm{c}=0$ where $\mathrm{g}$ is variable and $\mathrm{c}$ is a constant

The common radical axis is the $\mathrm{y}$-axis (centre on $\mathrm{x}$-axis)

The equation of system of co-axial circles in the simplest form is $x^2+y^2+2fy+c=0$ where fis a variable and $\mathrm{c}$ is a constant

The common radical axis is the $\mathrm{x}$-axis (centre on $\mathrm{y}$ axis)

1. Find the equation of tangent to the circle $x^2+y^2-2ax=0$ at the point $(a(1+\cos \alpha ),a\sin \alpha )$

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Solution:

Equation of tangent of $x^2+y^2-2ax=0at(a(1+\cos \alpha )$, asin $\alpha )$ is

$\mathrm{ax}(1+\cos \alpha )+\mathrm{aysin}\alpha -\mathrm{a}(\mathrm{x}+\mathrm{a}(1+\cos \alpha ))=0$

$\mathrm{ax}+\mathrm{axcos}\alpha +\mathrm{aysin}\alpha -\mathrm{ax}-{\mathrm{a}}^2(1+\cos \alpha )=0$

$\mathrm{axcos}\alpha +\mathrm{aysin}\alpha ={\mathrm{a}}^2(1+\cos \alpha )$

$\mathrm{x}\cos \alpha +\mathrm{y}\sin \alpha =\mathrm{a}(1+\cos \alpha )$

2. Find the equations of the tangents to the circle $x^2+y^2=9$ which make an angle of ${60}^{\circ }$ with the axis.

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Solution:

Since tangents make an angle of ${60}^{\circ }$ with the $\mathrm{x}$-axis so slope of tangent

$\mathrm{m}=\tan {60}^{\circ }=\mathrm{m}=\sqrt{3}$

radius of circle $x^2+y^2=9$ is 3

we know equation of tangent to a circle $x^2+y^2=a^2$ is

$\mathrm{y}=\mathrm{mx}\pm \mathrm{a}\sqrt{1+{\mathrm{m}}^2}$

$\therefore \mathrm{y}=\sqrt{3}\mathrm{x}\pm 3\sqrt{1+3}$

$=\sqrt{3}\mathrm{x}\pm 6$

or $\sqrt{3}x-y\pm 6=0$ ie., $\sqrt{3}x-y+6=0$ and $\sqrt{3}x-y-6=0$ are equations of tangents.

3. Show that the line $(x-2)\cos \theta +(y-2)\sin \theta =1$ touches a circle for all values of $\theta$. Find the circle.

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Solution:

Since the line $(\mathrm{x}-2)\cos \theta +(\mathrm{y}-2)\sin \theta =1$____________(1) touches a circle so it is a tangent equation to a circle.

Equation of tangent to a circle at $({\mathrm{x}}_1,{\mathrm{y}}_1)$ is $(\mathrm{x}-\mathrm{h}){\mathrm{x}}_1+(\mathrm{y}-\mathrm{k}){\mathrm{y}}_1={\mathrm{a}}^2$ to a circle $(\mathrm{x}-\mathrm{y}{)}^2+(\mathrm{y}-\mathrm{k}{)}^2={\mathrm{a}}^2$ comparing (1) and (2) we get

$\mathrm{x}-\mathrm{h}=\mathrm{x}-2$ $\mathrm{y}-\mathrm{k}=\mathrm{y}-2$ $$ and ${\mathrm{a}}^2=1$

${\mathrm{x}}_1=1\cos \theta$ ${\mathrm{y}}_1=1\sin \theta$

$\therefore$ Required equation of circle is

$(\mathrm{x}-2{)}^2+(\mathrm{y}-2{)}^2=1$

${\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-4\mathrm{y}+7=0$

4. Find the equation of the normal to the circle $x^2+y^2=2x$ which is parallel to the line $x+2y=3$

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Solution:

Equation of normal at $({\mathrm{x}}_1,{\mathrm{y}}_1)$ of ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}=0$ is

${\displaystyle \frac{\mathrm{x}-{\mathrm{x}}_1}{{\mathrm{x}}_1-1}}={\displaystyle \frac{\mathrm{y}-{\mathrm{y}}_1}{{\mathrm{y}}_1-0}}(\because {\displaystyle \frac{\mathrm{x}-{\mathrm{x}}_1}{{\mathrm{ax}}_1+\mathrm{g}}}={\displaystyle \frac{\mathrm{y}-{\mathrm{y}}_1}{{\mathrm{by}}_1+\mathrm{f}}})$

Slope of this equation is ${\displaystyle \frac{y_1}{x_1-1}}$

Slope of $x+2y=3$ is ${\displaystyle \frac{-1}{2}}$

Since given that normal is parallel to $x+2y=3$

$\therefore {\displaystyle \frac{{\mathrm{y}}_1}{{\mathrm{x}}_1-1}}={\displaystyle \frac{-1}{2}}$

$2{\mathrm{y}}_1={\mathrm{x}}_1+1$ therefore locus of $({\mathrm{x}}_1,{\mathrm{y}}_1)$ is ${\mathrm{x}}_1+2{\mathrm{y}}_1=1$

It is the equation of normal

5. Show that the line $3x-4y=1$ touches the circle $x^2+y^2-2x+4y+1=0$ find the coordinates of the point of contact.

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Solution:

Centre and radius of circle ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}+4\mathrm{y}+1=0$

is $(1,-2)$ and $\sqrt{1+4-1}=2$ respectively.

If the distance of a line $3x-4y=1$ from the centre $(1,-2)$ is equal to radius then the line touches or it is tangent to a circle.

$=\left|{\displaystyle \frac{3\times 1-4(-2)-1}{\sqrt{3^2+4^2}}}\right|(\left|{\displaystyle \frac{{\mathrm{ax}}_1+{\mathrm{by}}_1+\mathrm{c}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}}\right|=\mathrm{d})$

$=\left|{\displaystyle \frac{3+8-1}{5}}\right|$

$=2$

$\therefore$ line $3x-4y=1$ touches the circle.

Let point of contact be $(x_1,y_1)$ then equation of tangent to a circle $x^2+y^2-2x+4y+1=0$ is ${\mathrm{xx}}_1+{\mathrm{yy}}_1-(\mathrm{x}+{\mathrm{x}}_1)+2\left({\mathrm{y}}^{+}{\mathrm{y}}_1\right)+1=0$

$\mathrm{x}({\mathrm{x}}_1-1)+\mathrm{y}({\mathrm{y}}_1+2)-{\mathrm{x}}_1+2{\mathrm{y}}_1+1=0$____________________(1)

and given line $3x-4y-1=0$___________________(2)

(1) and (2) are idenfical then comparing (1) and (2) we get

${\displaystyle \frac{{\mathrm{x}}_1-1}{3}}={\displaystyle \frac{{\mathrm{y}}_1+2}{-4}}={\displaystyle \frac{-{\mathrm{x}}_1+2{\mathrm{y}}_1+1}{-1}}$

$-{\mathrm{x}}_1+1=-3{\mathrm{x}}_1+6{\mathrm{y}}_1+3$ or $2{\mathrm{x}}_1-6{\mathrm{y}}_1-2=0$

$-y_1-2=4x_1-8y_1-4$ or $4x_1-7y_1-2=0$

Solving these two equations of ${\mathrm{x}}_1,{\mathrm{y}}_1$ we get ${\mathrm{x}}_1={\displaystyle \frac{-1}{5}}$ and ${\mathrm{y}}_1={\displaystyle \frac{-2}{5}}$

$\therefore$ point of contact is $({\displaystyle \frac{-1}{5}},{\displaystyle \frac{-2}{5}})$

6. The angle between a pair of tangents from a point $P$ to the circle ${\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}-6\mathrm{y}+9{\sin }^2\alpha +13{\cos }^2\alpha =0$ is $2\alpha$. Find the equation of the locus of the point $P$.

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Solution:

Let the coordinate of $\mathrm{Pbe}({\mathrm{x}}_1,{\mathrm{y}}_1)$ and given

circle is ${\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}-6\mathrm{y}+9{\sin }^2\alpha +13{\cos }^2\alpha =0$

$(\mathrm{x}+2{)}^2+(\mathrm{y}-3{)}^2-4-9+9{\sin }^2\alpha +13{\cos }^2\alpha =0$

$(x+2{)}^2+(y-3{)}^2+9{\sin }^2\alpha -13(1-{\cos }^2\alpha )=0$

$(\mathrm{x}+2{)}^2+(\mathrm{y}-3{)}^2+9{\sin }^2\alpha -13{\sin }^2\alpha =0$

$(\mathrm{x}+2{)}^2+(\mathrm{y}-3{)}^2=4{\sin }^2\alpha =(2\sin \alpha {)}^2$

$\therefore$ centre and radius of circle is $(-2,3)$ and $2\sin \alpha$ respectively

Distance of $\mathrm{P}$ and $\mathrm{C}$ is

$\mathrm{PC}=\sqrt{{({\mathrm{x}}_1+2)}^2+{({\mathrm{y}}_1-3)}^2}$

In ${}_{\mathrm{\Delta }}\mathrm{PCR}$

$\sin \alpha ={\displaystyle \frac{2\sin \alpha }{\sqrt{{({\mathrm{x}}_1+2)}^2+{({\mathrm{y}}_1-3)}^2}}}$

or $\sqrt{{({\mathrm{x}}_1+2)}^2+{({\mathrm{y}}_1-3)}^2}=2$

Squaring

${({\mathrm{x}}_1+2)}^2+{({\mathrm{y}}_1-3)}^2=4$ or $(\mathrm{x}+2{)}^2+(\mathrm{y}-3{)}^2=4$

$\therefore$ locus of point $\mathrm{P}$ is a circle

7. If the length of tangent from $(f,g)$ to the circle $x^2+y^2=6$ be twice the length of the tangent from $(f,g)$ to circle ${\mathrm{x}}^2+{\mathrm{y}}^2+3\mathrm{x}+3\mathrm{y}=0$ then prove that ${\mathrm{f}}^2+{\mathrm{g}}^2+4\mathrm{f}+4\mathrm{g}+2=0$

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Solution:

According to the question

$\sqrt{f^2+g^2-6}=2\sqrt{f^2+g^2+3f+3g}$

Squaring both side

${\mathrm{f}}^2+{\mathrm{g}}^2-6=4({\mathrm{f}}^2+{\mathrm{g}}^2+3\mathrm{f}+3\mathrm{g})$

$3{\mathrm{f}}^2+3{\mathrm{g}}^2+12\mathrm{f}+12\mathrm{g}+6=0$

Divide by 3 we get ${\mathrm{f}}^2+{\mathrm{g}}^2+4\mathrm{f}+4\mathrm{g}+2=0$

8. The chord of contact of tangents drawn from a point on the circle $x^2+y^2=a^2$ to the circle $x^2+y^2=$ ${\mathrm{b}}^2$ touches the circle ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{c}}^2$ show that a.b.c are in GP.

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Solution:

Let $\mathrm{P}(\mathrm{a}\cos \theta ,\mathrm{asin}\theta )$ be a point on the circle ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2$ Then equation of chord of contact to the circle $x^2+y^2=b^2$ from $P(a\cos \theta ,a\sin \theta )$ is

$\mathrm{x}(\mathrm{a}\cos \theta )+\mathrm{y}(\mathrm{a}\sin \theta )={\mathrm{b}}^2$

$\mathrm{axcos}\theta +\mathrm{aysin}\theta ={\mathrm{b}}^2$

It is a tangent to the circle ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{c}}^2$

$\therefore$ length of perpendicular to the line $=$ radius.

$\left|{\displaystyle \frac{-b^2}{\sqrt{a^2}}}\right|=c$

${\mathrm{b}}^2=\mathrm{ac}$

$\therefore$ a, b.c are in G.P.

EXERCISE

1. The points $A$ and $B$ in a plane are such that for all points $P$ lies on circle Satisfying ${\displaystyle \frac{PA}{PB}}=k$, then $k$ will not be equal to

(a) 0

(b) 1

(c) 2

(d) None of these

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Answer: b

2. If the line $hx+ky=1$ touches $x^2+y^2=a^2$ then the locus of the point $(h,k)$ is a circle of radius

(a) a

(b) $1/\mathrm{a}$

(c) $\sqrt{\mathrm{a}}$

(d) ${\displaystyle \frac{1}{\sqrt{\mathrm{a}}}}$

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Answer: b

3. Equation of incircle of equilateral triangle $\mathrm{ABC}$ where $\mathrm{B}(2,0),\mathrm{C}(4,0)$ and $\mathrm{A}$ lies in the fourth quadrant is

(a) $x^2+y^2-6x+{\displaystyle \frac{2y}{\sqrt{3}}}+9=0$

(b) $x^2+y^2-6x-{\displaystyle \frac{2y}{\sqrt{3}}}+9=0$

(c) ${\mathrm{x}}^2+{\mathrm{y}}^2+6\mathrm{x}+{\displaystyle \frac{2\mathrm{y}}{\sqrt{3}}}+9=0$

(d) None of these

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Answer: a

4. A variable circle having radius ’ $a$ ’ passes through origin meets the coordinate axes in points A and B. locus of centroid of triangle $\mathrm{OAB}$, ’ $\mathrm{O}$ ’ being the origin is

(a) $9(x^2+y^2)=4a^2$

(b) $9(x^2+y^2)=a^2$

(c) $9(x^2+y^2)=2a^2$

(d) $9(x^2+y^2)=8a^2$

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Answer: a

5. The locus of the centre of the circle which cuts a chord of length 2 a from the positive $x$-axis and passes through a point on positive $y$-axis distant $b$ from the origin is

(a) $x^2+2by=b^2+a^2$

(b) $x-2by=b^2+a^2$

(c) $x+2by=a^2-b^2$

(d) $x^2-2by=b^2-a^2$

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Answer: c

6. The number of circle having radius 5 and passing through the points $(-2,0)$ and $(4,0)$ is

(a) one

(b) two

(c) four

(d) infinite

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Answer: b

7. The equation of the smallest circle passing through the intersection of the line $x+y=1$ and the circle ${\mathrm{x}}^2+{\mathrm{y}}^2=9$ is

(a) ${\mathrm{x}}^2+{\mathrm{y}}^2+\mathrm{x}+\mathrm{y}-8=0$

(b) $x^2+y^2-x-y-8=0$

(c) $x^2+y^2-x-y+8=0$

(d) None of these

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Answer: b

8. The number of the points on the circle $x^2+y^2-4x-10y+13=0$ which are at a distance 1 from the point $(-3,2)$ is

(a) 1

(b) 2

(c) 3

(d) None of these

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Answer: d

9. The locus of the mid-point of a chord of the circle $x^2+y^2=4$ which subtends a right angle at the origin is

(a) $x+y=2$

(b) $x^2+y^2=1$

(c) $x^2+y^2=2$

(d) $x+y=1$

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Answer: b

10. The area of the triangle formed by joining the origin to the points of intersection of the line $x\sqrt{5}+2y$

$=3\sqrt{5}$ and circle ${\mathrm{x}}^2+{\mathrm{y}}^2=10$ is

(a) 3

(b) 4

(c) 5

(d) 6

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Answer: c

11. If $(-3,2)$ lies on the circle $x^2+y^2+2gx+2fy+c=0$ which is concentric with the circle $x^2+y^2+6x+8y-$ $5=0$ then $\mathrm{c}$ is

(a) 11

(b) -11

(c) 24

(d) None of these

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Answer: b

12. A variable circle passes through the fixed point A(p.q) and touches $\mathrm{x}$-axis. The locus of the other end of the diameter through $A$ is

(a) $(x-p{)}^2=4qy$

(b) $(x-q{)}^2=4qy$

(c) $(\mathrm{y}-\mathrm{p}{)}^2=4\mathrm{qx}$

(d) $(\mathrm{y}-\mathrm{q}{)}^2=4\mathrm{px}$

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Answer: a

13. If the lines $2x+3y+1=0$ and $3x-y-4=0$ lie along diameters of a circle of circumference $10\pi$, then the equation of the circle is

(a) $x^2+y^2-2x+2y-23=0$

(b) $x^2+y^2+2x+2y-23=0$

(c) $x^2+y^2-2x-2y-23=0$

(d) $x^2+y^2+2x-2y-23=0$

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Answer: a

14. The lines $2x-3y=5$ and $3x-4y=7$ are diameters of a circle having area as 154 sq.unit, then the equation of the circle is

(a) $x^2+y^2+2x-2y=62$

(b) $x^2+y^2+2x-2y=47$

(c) $x^2+y^2-2x+2y=47$

(d) $x^2+y^2-2x+2y=62$

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Answer: c

15. The equation of circle which passes through the origin and cuts off intercepts 5 and 6 from the positive parts of the axes respectively, is

(a) $(x+5/2{)}^2+(y+3)={\displaystyle \frac{61}{4}}$

(b) $(x-5/2{)}^2+(y-3{)}^2={\displaystyle \frac{61}{4}}$

(c) $(x-5/2{)}^2-(y-3{)}^2={\displaystyle \frac{61}{4}}$

(d) $(x-5/2{)}^2+(y+3{)}^2={\displaystyle \frac{61}{4}}$

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Answer: b