Topics covered :

1. Distance of a line from a circle

2. Different forms of the equations of tangents

3. Length of tangent from a point to a circle.

4. Normal to a circle at a given point.

1. 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

2. Equations of tangents

1. 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 : ${\mathrm{xx}}_1+{\mathrm{yy}}_1={\mathrm{a}}^2$

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 ${\mathrm{xx}}_1+{\mathrm{yy}}_1+\mathrm{g}(\mathrm{x}+{\mathrm{x}}_1)+f(\mathrm{y}+{\mathrm{y}}_1)+\mathrm{c}=0$

I. Parametric forms

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

II. 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 $(\pm \frac{\mathrm{am}}{\sqrt{1+{\mathrm{m}}^2}},\mp \frac{\mathrm{a}}{\sqrt{1+{\mathrm{m}}^2}})$

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

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

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

(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 $(\frac{-m{r}^2}{c},\frac{r^2}{c})$

(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 $(\frac{-{\mathrm{ar}}^2}{\mathrm{c}},\frac{-{\mathrm{br}}^2}{\mathrm{c}})$

III. 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

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

$\mathrm{P}({\mathrm{x}}_1,{\mathrm{y}}_1)$ 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

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

${\mathrm{m}}^2({\mathrm{x}}_1{}^2-{\mathrm{a}}^2)-2{\mathrm{mx}}_1{\mathrm{y}}_1+{\mathrm{y}}_1{}^2-{\mathrm{a}}^2=0$ is a quadratic equation

in $m$ which gives 2 values of $m$.

We get two equations of tangents.

Tangents from a point outside the circle

Let the circle be ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{r}}^2$ and a point $\mathrm{P}\left({\mathrm{x}}_{1,}{\mathrm{y}}_1\right)$ 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 $\mathrm{m}$ solve and get two values of $m$ and substitute in (1)

IV. Length of the tangent from a point to a circle.

Let the circle be $S_{\equiv x^2+y^2+2gx+2fy+c}=0$ then centre and radius of circle are $(-\mathrm{g},-f)$ and $\sqrt{g^2+f^2-c}$ respectively and $\mathrm{let}\mathrm{P}({\mathrm{x}}_1,{\mathrm{y}}_1)$ be any point outside the circle.

$\mathrm{PT}=\sqrt{x_1^2+y_1^2+2g{x}_1+2f_1+y_1+c}$

$=\sqrt{S_1}$

V. 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 ${\mathrm{x}}_1{}^2+{\mathrm{y}}_1{}^2+2{\mathrm{gx}}_1+2f{\mathrm{y}}_1+\mathrm{c}=0\Rightarrow$ ${\mathrm{S}}_1=0(:\mathrm{PA}\cdot \mathrm{PB}=\mathrm{PC}\cdot \mathrm{PD}={\mathrm{PT}}^2)$

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

VI. 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 $S=x^2+y^2+2gx+2fy+c$

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

Note: The pair of tangents from $(0,0)$ to the circle

VII. 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.

(1) Point form

To find the equation of normal to the circle ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2$ 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 $\frac{\mathrm{y}-0}{{\mathrm{y}}_1-0}=\frac{\mathrm{x}-0}{{\mathrm{x}}_1-0}$

or

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

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

To find normal at $(x_1,y_1)$ of second degree conics $a^2+2hxy+b{y}^2+2gx+2fy+c=0\dots \dots \dots .(1)$

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

Write first two rows as ax ${}_1+{\mathrm{hy}}_1+\mathrm{g}$ and ${\mathrm{hx}}_1+{\mathrm{by}}_1+\mathrm{f}$

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

• If equation of circle is ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2$

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

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

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

$$$\frac{x}{x_1}=\frac{y}{y_1}$ is equation of normal at $(x_1,y_1)$

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

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

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