1. 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} $

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

2. Intersection of two circles, common Tangents to two circles

Let the two circles be $\left(x-g _{1}\right)^{2}+\left(y-f _{1}\right)^{2}=r _{1}{ }^{2}$ and $\left(x-g _{2}\right)^{2}+\left(y-f _{2}\right)^{2}=r _{2}{ }^{2}$

with centres $\mathrm{C} _{1}\left(\mathrm{~g} _{1}, \mathrm{f} _{1}\right)$ and $\mathrm{C} _{2}\left(\mathrm{~g} _{2}, \mathrm{f} _{2}\right)$ 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

$\frac{\mathrm{C}_{1} \mathrm{E}}{\mathrm{C} _{2} \mathrm{E}}=\frac{\mathrm{r} _{1}}{\mathrm{r} _{2}} \& \frac{\mathrm{C} _{1} \mathrm{~F}}{\mathrm{C} _{2} \mathrm{~F}}=\frac{\mathrm{r} _{1}}{\mathrm{r} _{2}}$

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 $\mathrm{m}$. Substitute the values of $m$ we get required respective common tangent equation

Case II When $\mid \mathrm{C} _{1} \mathrm{C} _{2} \mathrm{I}=\mathrm{r} _{1}+\mathrm{r} _{2}$

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 $\left(\frac{r _{1} g _{2}+r _{2} g _{1}}{r _{1}+r _{2}}, \frac{r _{1} f _{2}+r _{2} f _{1}}{r _{1}+r _{2}}\right)$

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.

Case IV When $\left|C _{1} C _{2}\right|=\left|r _{1}-r _{2}\right|$

Then the two circles touch each other internally.

In this case only one direct common tangent.

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 $\mathrm{F}$ are $\left(\frac{\mathrm{r} _{1} \mathrm{~g} _{2}-\mathrm{r} _{2} \mathrm{~g} _{1}}{\mathrm{r} _{1}-\mathrm{r} _{2}}, \frac{\mathrm{r} _{1} \mathrm{f} _{2}-\mathrm{r} _{2} \mathrm{f} _{1}}{\mathrm{r} _{1}-\mathrm{r} _{2}}\right)$

Case V When $\left|\mathrm{C} _{1} \mathrm{C} _{2}\right|<\left|\mathrm{r} _{1}-\mathrm{r} _{2}\right|$

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.

3. Length of direct common tangent if $\left|\mathrm{C} _{1} \mathbf{C} _{2}\right| \mathbf{r} _{1}+\mathbf{r} _{2}$

length of direct common tangent $=\sqrt{\mathrm{d}^{2}-\left(\mathrm{r} _{1}-\mathrm{r} _{2}\right)^{2}}$

length of transverse common tangent $=\sqrt{\mathrm{d}^{2}-\left(\mathrm{r} _{1}+\mathrm{r} _{2}\right)^{2}}$

Where $\mathrm{d}=\left|\mathrm{C} _{1} \mathrm{C} _{2}\right|$ and $\mathrm{r} _{1}, \mathrm{r} _{2}$ are radii of the circles

4. Pole and Polar of the circle

Let $\mathrm{P}\left(\mathrm{x} _{1}, \mathrm{y} _{1}\right)$ be any point inside or outside the circle. Draw chords $\mathrm{AB}$ and $\mathrm{CD}$ passing through $\mathrm{P}$ . If tangents to the circle at $\mathrm{A}$ and $\mathrm{B}$ meet at $\mathrm{Q}(\mathrm{h}, \mathrm{k})$ then locus of $\mathrm{Q}$ is called the polar of $\mathrm{P}$ with respect to circle and $\mathrm{P}$ is called the pole and if tangents to the circle at $\mathrm{C}$ and $\mathrm{D}$ meet at $\mathrm{R}$, then the straight line $\mathrm{QR}$ is polar with $\mathrm{P}$ as its pole.

Let the equation of circle be $\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{a}^{2}$ then $\mathrm{AB}$ is a chord of contact of $\mathrm{Q}(\mathrm{h}, \mathrm{k})$

$\therefore \mathrm{xh}+\mathrm{yk}=\mathrm{a}^{2}$ is its equation

$\mathrm{P}\left(\mathrm{x} _{1}, \mathrm{y} _{1}\right)$ lies on $\mathrm{AB}$

$\therefore \mathrm{hx} _{1}+\mathrm{ky} _{1}=\mathrm{a}^{2}$.

Hence locus of $\mathrm{Q}(\mathrm{h}, \mathrm{k})$ is $\mathrm{xx} _{1}+\mathrm{yy} _{1}=\mathrm{a}^{2}$ which is polar of $\mathrm{P}\left(\mathrm{x} _{1}, \mathrm{y} _{1}\right)$ with respect to circle $\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{a}^{2}$ Equation of polar of the circle $x^{2}+y^{2}+2 g x+2 f y+c=0$ with respect to $\left(x _{1}, y _{1}\right)$ is

$ \mathrm{xx} _{1}+\mathrm{yy} _{1}+\mathrm{g}\left(\mathrm{x}+\mathrm{x} _{1}\right)+\mathrm{f}\left(\mathrm{y}+\mathrm{y} _{1}\right)+\mathrm{c}=0 $

Coordinates of pole of a line

Let the polar line $a x+b y+c=0$ with respect to the circle $x^{2}+y^{2}=r^{2}$

Let the pole be $\left(\mathrm{x} _{1}, \mathrm{y} _{1}\right)$ then equation of polar with respect to the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{r}^{2}$ is

$\mathrm{xx} _{1}+\mathrm{yy} _{1}-\mathrm{r}^{2}=0 ……….(1)$

Which is same as ax $+b y+c=0………..(2)$

comparing (1) and (2) we get

$ \begin{aligned} & \frac{x _{1}}{a}=\frac{y _{1}}{b}=\frac{-r^{2}}{c} \\ & x _{1}=\frac{-a^{2}}{c} \& y _{1}=\frac{-b r^{2}}{c} \end{aligned} $

$\therefore$ The coordinate of pole is $\left(\frac{-\mathrm{ar}^{2}}{\mathrm{c}}, \frac{-\mathrm{br}^{2}}{\mathrm{c}}\right)$

Properties of pole and polar

(i) The distance of any two points $\mathrm{P}\left(\mathrm{x} _{1}, \mathrm{y} _{1}\right)$ and $\mathrm{Q}\left(\mathrm{x} _{2}, \mathrm{y} _{2}\right)$ from the centre of a circle is proportional to the distance of each from the polar to the other

(ii) If $\mathrm{O}$ be the centre of a circle and $\mathrm{P}$ any point then $\mathrm{OP}$ is perpendicular to the polar of $\mathrm{P}$

(iii) If $\mathrm{O}$ be the centre of a circle and $\mathrm{P}$ any point then if $\mathrm{OP}$ (produced if necessary) meet the polar of $\mathrm{P}$ in $\mathrm{S}$ then OP.OS $=\mathrm{r}^{2}$

(iv) If the polar of $\mathrm{P}\left(\mathrm{x} _{1}, \mathrm{y} _{1}\right)$ with respect to a circle passes through $\mathrm{S}\left(\mathrm{x} _{2}, \mathrm{y} _{2}\right)$ then the ploar of $\mathrm{S}$ will pass through $P$ and such points are said to be conjugate points

(v) If the pole of the line $a x+b y+c=0$ with respect to a circle lies on another line $a _{1} x+b _{1} y+c _{1}=0$. Then the pole of the second line will lie on the first line and such lines are said to be conjugate lines.

Note:

(i) $\quad \mathrm{P}\left(\mathrm{x} _{1}, \mathrm{y} _{1}\right)$ and $\mathrm{Q}\left(\mathrm{x} _{2}, \mathrm{y} _{2}\right)$ are convent points w.r.t the circle $\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0$ if $\mathrm{x} _{1} \mathrm{x} _{2}+\mathrm{y} _{1} \mathrm{y} _{2}+\mathrm{g}\left(\mathrm{x} _{1}+\mathrm{x} _{2}\right)+\mathrm{f}\left(\mathrm{y} _{1}+\mathrm{y} _{2}\right)+\mathrm{c}=0$

(ii) If $\mathrm{P}$ and $\mathrm{Q}$ are confined points w.r.t to a circle with centre at $\mathrm{O}$ and radius $\mathrm{r}$, then $\mathrm{PQ}^{2}=\mathrm{OP}^{2}+\mathrm{OQ}^{2}-2 \mathrm{r}^{2}$.

5. Common Chord of two circles

The common chord joining the point of intersection of two circles is called their common chord .If $\mathrm{S}=0$ and $\mathrm{S}^{1}=0$ be two interesting circles then the equation of their common chord is $\mathrm{S}-\mathrm{S}^{1}=0$ Let the equations of circle one

$S=x^{2}+y^{2}+2 g _{1} x+2 f _{1} y+c _{1}=0$

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

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

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

$2 x\left(g _{1}-g _{2}\right)+2 y\left(f _{1}-f _{2}\right)+c _{1}-c _{2}=0$

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.

Exercises

1. There are two circles whose equations are $x^{2}+y^{2}=9$ and $x^{2}+y^{2}-8 x-6 y+n^{2}=O n \in Z$. If two circles have exactly two common tangents then the number of possible values of $n$

2. Find all the common tangents to the circles $x^{2}+y^{2}-2 x-6 y+9=0$ and $x^{2}+y^{2}+6 x-2 y+1=0$

Show Answer

Solution : Centre and radius of circle $x^{2}+y^{2}-2 x-6 y+9=0$ is $C _{1}(1,3)$ and $r _{1}=1$ respectively Centre and radius of circle $x^{2}+y^{2}+6 x-2 y+1=0$ is $C _{2}(-3,1)$ and $r _{2}=3$

$ \begin{aligned} \mathrm{C} _{2} \mathrm{C} _{2} & =\sqrt{4^{2}+2^{2}} & & \mathrm{r} _{1}+\mathrm{r} _{2}=1+3=4 \\ & =\sqrt{16+4} & & \therefore\left|\mathrm{C} _{1} \mathrm{C} _{2}\right|>\mathrm{r} _{1}+\mathrm{r} _{2} \\ & =\sqrt{20} & & \therefore \text { These are four common tangents } \\ & =2 \sqrt{5} & & 2 \text { direct and } 2 \text { transverse. } \end{aligned} $

To find equations of direct common tangents

Since the coordinate of $E$ divides the line $\mathrm{C} _{1} \mathrm{C} _{2}$ in the ratio $\mathrm{r} _{1}: \mathrm{r} _{2}$ ie., $1: 3$ externally

$\therefore$ coordinate of $\mathrm{E}$ is

$\left(\frac{1(-3)-3.1}{1-3}, \frac{1.1-3.3}{1-3}\right)$ or $E(3,4)$

Equation of $\mathrm{AB}$ through $\mathrm{E}(3,4)$ with slope $\mathrm{m}$ is $\mathrm{y}-4=\mathrm{m} _{1}(\mathrm{x}-3)$ or $\mathrm{m} _{1} \mathrm{x}-\mathrm{y}+\left(4-3 \mathrm{~m} _{1}\right)=0$

If is a tangent to the circle so distance of the line from the centre is equal to radius

$ \begin{aligned} \therefore \quad 3 & =\left|\frac{-3 \mathrm{~m} _{1}-1+4-3 \mathrm{~m} _{1}}{\sqrt{\mathrm{m} _{1}^{2}+1}}\right| \\ 3 & =\left|\frac{3-6 \mathrm{~m} _{1}}{\sqrt{\mathrm{m} _{1}^{2}+1}}\right| \\ 3 & =\frac{3-6 \mathrm{~m} _{1}}{\sqrt{\mathrm{m} _{1}^{2}+1}} \\ & \sqrt{\mathrm{m} _{1}^{2}+1} \mid=1-2 \mathrm{~m} _{1} \end{aligned} $

Squaring

$\mathrm{m} _{1}^{2}+1=1+4 \mathrm{~m} _{1}^{2}-4 \mathrm{~m} _{1}$ $3 \mathrm{~m} _{1}{ }^{2}-4 \mathrm{~m} _{1}=0 \Rightarrow \mathrm{m} _{1}=0$ or $\mathrm{m} _{1}=4 / 3$

Substitute the values of $m _{1}$ in equation (1) we get the equations of tangents $y=4$ and $4 x-3 y=0$

To find the indirect or transverse common tangents

The coordinates of $\mathrm{T}$ dividing

$\mathrm{C} _{1} \mathrm{C} _{2}$ internally in the ratio

$\mathrm{r} _{1}: \mathrm{r} _{2}=1: 3$ is

$\left(\frac{-3+3}{4}, \frac{1+9}{4}\right)=\left(0, \frac{5}{2}\right)$

Equation of line through $\mathrm{T}$ with slope $\mathrm{m} _{2}$ is

$y-5 / 2=m _{2}(x-0)$ or $2 m _{2} x-2 y+5=0$

Since it is a tangent to the circles

$\therefore$ distance of the line from its centre is equal to radius.

$ \begin{aligned} & \therefore \quad\left|\frac{2 \mathrm{~m} _{2}(-3)-2(1)+5}{\sqrt{4 \mathrm{~m} _{2}^{2}+4}}\right|=3 \\ & \left|\frac{-6 m _{2}+3}{2 \sqrt{m _{2}^{2}+1}}\right|=3 \\ & \text { or } 3-6 \mathrm{~m} _{2}=6 \sqrt{\mathrm{m} _{2}^{2}+1} \\ & 1-2 m _{2}=2 \sqrt{m _{2}^{2}+1} \\ & \text { Squaring } \\ & 1+4 m _{2}^{2}-4 m _{2}=4 m _{2}^{2}+4 \\ & 4 m _{2}=-3 \text { or } m _{2}=\frac{3}{4} \text { and } \infty\left(\text { as coefficient of } m _{2}^{2}=0\right) \end{aligned} $

$\therefore$ The equations of transverse common tangents are

$ 3x+4 y-10=0 \text { and } x=0 $

3. If the circle $\mathrm{C} _{1}=\mathrm{x}^{2}+\mathrm{y}^{2}=16$ intersects another circles $\mathrm{C} _{2}$ of radius 5 in such a manner that the common chord is of maximum length and has a slope equal to $3 / 4$ the find the coordinates of the centre $\mathrm{C} _{2}$

Show Answer

Solution: $\left(\frac{9}{5}, \frac{-12}{5}\right) \& \left(\frac{-9}{5}, \frac{12}{5}\right)$

When two circles intersect, the common chord of maximum length will be the diameter of smaller circle.

$\therefore \mathrm{AB}$ is diameter of smaller circle

$\mathrm{C} _{2} \mathrm{~A}=5$ and $\mathrm{C} _{1} \mathrm{~A}=4$

$\therefore \mathrm{C} _{1} \mathrm{C} _{2}=\sqrt{5^{2}-4^{2}}$

$\mathrm{C} _{1} \mathrm{C} _{2}=3$

Given that slope of $A B$ is $3 / 4$

$\therefore$ slope of $\mathrm{C} _{1} \mathrm{C} _{2}$ is $-4 / 3$

and slope of $\mathrm{C} _{1} \mathrm{C} _{2}$ is $\frac{\mathrm{k}}{\mathrm{h}}$

$ \begin{aligned} & \therefore \frac{-4}{3}=\frac{\mathrm{k}}{\mathrm{h}} \\ & \mathrm{k}=\frac{-4}{3} \mathrm{~h} \end{aligned} $

$\mathrm{C} _{1} \mathrm{C} _{2}=\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}=3$

or $\mathrm{h}^{2}+\mathrm{k}^{2}=9$

Substitute $\mathrm{k}=\frac{-4}{3} \mathrm{~h}$ in equation $(1)$

$\mathrm{h}^{2}+\frac{16}{9} \mathrm{~h}^{2}=9$

$25 \mathrm{~h}^{2}=81$

$\mathrm{h}^{2}=\frac{81}{25}$

$\mathrm{h}= \pm \frac{9}{5} \quad \therefore \mathrm{K}= \pm \frac{12}{5}$

$\therefore$ coordinates of $\mathrm{C} _{2}$ are $\left(\frac{9}{5}, \frac{-12}{5}\right)$ and $\left(\frac{-9}{5}, \frac{12}{5}\right)$

4. The circle $x^{2}+y^{2}-4 x-8 y+16=0$ rolls up the tangent to it at $(2+\sqrt{3}, 3)$ by 2 units, assuming the $x-$ axis as horizontal, find the equation of the circle in the new position.

Show Answer

Solution:

Given circle is $x^{2}+y^{2}-4 x-8 y+16=0$

Equation of tangent to the circle at $(2+\sqrt{3}, 3)$ is $(2+\sqrt{3}) x+3 y-2(2+\sqrt{3}+x)-4(y+3)+16=0$ $\sqrt{3} \mathrm{x}-\mathrm{y}-2 \sqrt{3}=0$

slope of this line $\tan \theta=\sqrt{3} \quad \therefore \theta=60^{\circ}$.

$A$ and $B$ are the centres of

The circles in old and new positions respectively then

$\mathrm{A}(2,4)$ and $\left.\mathrm{B}\left(2+2 \cos 60^{\circ}\right), 4+2 \sin 60\right)(\Rightarrow \mathrm{AB}=2)$

$\therefore \mathrm{B}(3,4+\sqrt{3})$

Hence equation of circle at new position is

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

$\mathrm{x}^{2}+\mathrm{y}^{2}-6 \mathrm{x}-2(4+\sqrt{3}) \mathrm{y}+24+8 \sqrt{3}=0$

5. The pole of a straight line with respect to the circle $x^{2}+y^{2}=a^{2}$ lies on the circle $x^{2}+y^{2}=9 a^{2}$

Prove that the straight line touches the circle $x^{2}+y^{2}=\frac{a^{2}}{9}$

6. Prove that the polar of a given point with respect to any one of the circles $x^{2}+y^{2}-2 k x+c^{2}=0$, where $k$ is a variable, always passes through a fixed point, whatever be the value of $k$.

Practice questions

Passage 1

Let a straight line be drawn from a point $\mathrm{P}$ to meet the circle in $\mathrm{Q}$ and $\mathrm{R}$. Let the tangents at $\mathrm{Q}$ and $\mathrm{R}$ meet at $\mathrm{T}$. The locus of $\mathrm{T}$ is called the polar of $\mathrm{P}$ with respect to the circle. The given point $\mathrm{P}$ is called the pole of the polar line

Let $\mathrm{P}\left(\mathrm{x} _{1}, \mathrm{y} _{1}\right)$ be the given point lying outside the circle in fig (i) and inside the circle in fig (ii)

Through P, draw a line to meet the circle in Q and R. Let the tangents to the circle at Q and R meet in $\mathrm{T}(\mathrm{h}, \mathrm{k})$.

It is required to find the polar of Pie., locus of T.

Equation of $\mathrm{QR}$ the chord of contact of the tangents draw from $\mathrm{T}$ to the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{a}^{2}$ is $\mathrm{xh}+\mathrm{yk}$ $=\mathrm{a}^{2}…………. (1)$

since (1) passes through $\mathrm{P}\left(\mathrm{x} _{1}, \mathrm{y} _{1}\right)$

$\therefore \mathrm{x} _{1} \mathrm{~h}+\mathrm{y} _{1} \mathrm{k}=\mathrm{a}^{2}$

$\therefore$ The locus of $(\mathrm{h}, \mathrm{k})$ is $\mathrm{xx} _{1}+\mathrm{yy} _{1}=\mathrm{a}^{2}$, which is the equation of polar of $\mathrm{P}$.

1. If the polar of $\mathrm{P}$ with respect to the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{a}^{2}$ touches the circle $(\mathrm{x}-\mathrm{f})^{2}+(\mathrm{y}-\mathrm{g})^{2}=\mathrm{b}^{2}$ then its locus is given by the equation.

(a) $\left(f x+g y-a^{2}\right)^{2}=a^{2}\left(x^{2}+y^{2}\right)$

(c) $\left(\mathrm{fx}-\mathrm{gy}-\mathrm{a}^{2}\right)^{2}=\mathrm{a}^{2}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)$

(b) $\left(f x+g y-a^{2}\right)^{2}=b^{2}\left(x^{2}+y^{2}\right)$

(d) None of these.

2. The pole of the line $3 x+4 y=45$ with respect to the circle $x^{2}+y^{2}-6 x-8 y+5=0$ is

(a) $(6,8)$

(b) $(6,-8)$

(c) $(-6,8)$

(d) $(-6,-8)$

3. The pole of the chord of the circle $x^{2}+y^{2}=16$ which is bisected at the point $(-2,3)$ with respect to the circle is

(a) $\left(\frac{-32}{13}, \frac{48}{13}\right)$

(b) $\left(\frac{32}{13}, \frac{48}{13}\right)$

(c) $\left(\frac{-32}{13}, \frac{-48}{13}\right)$

(d) None of these

4. The coordinates of the poles of thre common chord of the circles $x^{2}+y^{2}=12$ and $\mathrm{x}^{2}+\mathrm{y}^{2}-5 \mathrm{x}+2 \mathrm{y}-2=0$ with respect to the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=12$ are

(a) $\left(6, \frac{-12}{5}\right)$

(b) $\left(-6, \frac{-12}{5}\right)$

(c) $\left(6, \frac{12}{5}\right)$

(d) None of these

5. The matching grid

6. If the line $x \cos \theta+y \sin \theta=2$ is the equation of a transverse common tangent to the circles $x^{2}+y^{2}$ $=4$ and $x^{2}+y^{2}-6 \sqrt{3} x-6 y+20=0$, Then the value of $\theta$ is

(a) $5 \pi / 6$

(b) $2 \pi / 3$

(c) $\pi / 3$

(d) $\pi / 6$

7. Two circles of radii $4 \mathrm{~cm}$ and $1 \mathrm{~cm}$ touch each other externally and $\theta$ is the angle contained by their direct common tangents. Then $\sin \theta$ is equal to

(a) $\frac{24}{25}$

(b) $\frac{12}{25}$

(c) $\frac{3}{4}$

(d) None of these.

Examples

1. If the tangent at the point $P$ on the circle $x^{2}+y^{2}+6 x+6 y=2$ meets the straight line $5 x-2 y+6=0$ at a point $Q$ on the $y$-axis, then the length of $P Q$ is

(a) 4

(b) $2 \sqrt{5}$

(c) 5

(d) $3 \sqrt{5}$

Show Answer

Solution. c

$Q(0,3)$ is a point where the line $5 x-2 y+6=0$ meets $y$-axis and the trangent drawn at $P$ $\mathrm{PQ}$ is a tangent

length of tangent $P Q=\sqrt{S} _{1}=\sqrt{0+9+0+6 \times 3-2}$

$ \begin{aligned} & =\sqrt{9+18-2} \\ & =\sqrt{25} \\ & =5 \end{aligned} $

2. If the two circles $x^{2}+y^{2}+2 g x+2 f y=0$ and $x^{2}+y^{2}+2 g _{1} x+2 f _{1} y=0$ and touch each other, then

(a) $f _{1} g=f g _{1}$

(b) $\mathrm{ff} _{1}=\mathrm{gg} _{1}$

(c) $\mathrm{f}^{2}+\mathrm{g}^{2}=\mathrm{f} _{1}^{2}+\mathrm{g} _{1}^{2}$

(d) None of these

Show Answer

Solution: (a)

Centres of the circles are $(-\mathrm{g}, \mathrm{f})$ and $\left(-\mathrm{g} _{1},-\mathrm{f} _{1}\right)$ and radius of the circles are $\sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}}$ and $\sqrt{\mathrm{g} _{1}{ }^{2}+\mathrm{f} _{1}{ }^{2}}$ If two circles touch each other then

$\left|\mathrm{C} _{1} \mathrm{C} _{2}\right|$ $=\left|\mathrm{r} _{1}+\mathrm{r} _{2}\right|($ external $)$

or $\left|r _{1}-r _{2}\right|$ (internal)

$\therefore \sqrt{\left(\mathrm{g}-\mathrm{g} _{1}\right)^{2}+\left(\mathrm{f}-\mathrm{f} _{1}\right)^{2}}=\sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}} \pm \sqrt{\mathrm{g} _{1}{ }^{2}+\mathrm{f} _{1}{ }^{2}}$

Squaring both sides

$\left(\mathrm{g}-\mathrm{g} _{1}\right)^{2}+\left(\mathrm{f}-\mathrm{f} _{1}\right)^{2}=\mathrm{g}^{2}+\mathrm{f}^{2}+\mathrm{g} _{1}{ }^{2}+\mathrm{f} _{1}{ }^{2} \pm 2 \sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}} \sqrt{\mathrm{g} _{1}{ }^{2}+\mathrm{f} _{1}{ }^{2}}$

$\mathrm{g}^{2}+\mathrm{g} _{1}{ }^{2}-2 \mathrm{gg} _{1}+\mathrm{f}^{2}+\mathrm{f} _{1}{ }^{2}-2 \mathrm{ff} _{1}=\mathrm{g}^{2}+\mathrm{f}^{2}+\mathrm{g} _{1}{ }^{2}+\mathrm{f} _{1}{ }^{2} \pm 2 \sqrt{\mathrm{g}^{2} \mathrm{~g} _{1}{ }^{2}+\mathrm{g}^{2} \mathrm{f} _{1}{ }^{2}+\mathrm{f}^{2} \mathrm{~g} _{1}{ }^{2}+\mathrm{f}^{2} \mathrm{f} _{1}{ }^{2}}$

$-\left(\mathrm{gg} _{1}+\mathrm{ff} _{1}\right)= \pm \sqrt{\mathrm{g}^{2} \mathrm{~g} _{1}{ }^{2}+\mathrm{g}^{2} \mathrm{f} _{1}{ }^{2}+\mathrm{f}^{2} \mathrm{~g} _{1}{ }^{2}+\mathrm{f}^{2} \mathrm{f} _{1}{ }^{2}}$

$\mathrm{g}^{2} \mathrm{~g} _{1}{ }^{2}+\mathrm{f}^{2} \mathrm{f} _{1}{ }^{2}+2 \mathrm{gg} _{1} \mathrm{ff} _{1}=\mathrm{g}^{2} \mathrm{~g} _{1}{ }^{2}+\mathrm{g} _{2} \mathrm{f} _{1}{ }^{2}+\mathrm{f}^{2} \mathrm{~g} _{1}{ }^{2}+\mathrm{f}^{2} \mathrm{f} _{1}{ }^{2}$

$\mathrm{g}^{2} \mathrm{f} _{1}{ }^{2}+\mathrm{f}^{2} \mathrm{~g} _{1}{ }^{2}-2 \mathrm{gfg} _{1} \mathrm{f} _{1}=0=\left(\mathrm{gf} _{1}-\mathrm{fg} _{1}\right)^{2}$

$\therefore \mathrm{gf} _{1}-\mathrm{fg} _{1}=0 \Rightarrow \mathrm{gf} _{1}=\mathrm{fg} _{1}$

3. The circles $(x-a)^{2}+(y-b)^{2}=c^{2}$ and $(x-b)^{2}+(y-a)^{2}=c^{2}$ touch each other, then

(a) $a=b \pm 2 c$

(b) $\mathrm{a}=\mathrm{b} \pm \sqrt{2} \mathrm{c}$

(c) $\mathrm{a}=\mathrm{b} \pm \mathrm{c}$

(d) None of these