Circle

Given, equation of circle

$ x^2+y^2-2 x-1=0 $

Coordinate of $C$ is $(1,0)$.

Radius $=\sqrt{2}$

Let the coordinate of $P$ be $(I+\sqrt{2} \cos \theta, \sqrt{2} \sin \theta)$.

Given equation of another circle

$ x^2+y^2-2 x=0 $

Coordinate center is $(1,0)$ and radius $=1$

∴ The two given circle are concentric.

Let the circumcenter of $C A B$ be $(h, k)$.

Circumcenter of $C A B$ lies on mid-point of $P C$

$ \begin{aligned} &\begin{array}{ll} & h=\frac{1+\sqrt{2} \cos \theta+1}{2} \\ \text { and } & k=\frac{\sqrt{2} \sin \theta+0}{2} \\ \Rightarrow & \cos \theta=\frac{2 h-2}{\sqrt{2}} \text { and } \sin \theta=\frac{2 k}{\sqrt{2}} \\ \Rightarrow & \cos ^2 \theta+\sin ^2 \theta=1 \\ \Rightarrow & \left(\frac{2(h-1)}{\sqrt{2}}\right)^2+\left(\frac{2 k}{\sqrt{2}}\right)^2=1 \\ \Rightarrow & 2(h-1)^2+2 k^2=1 \\ \Rightarrow & 2 h^2+2-4 h+2 k^2=1 \\ \Rightarrow & 2 h^2+2 k^2-4 h+1=0 \end{array}\\ &\text { ∴ The locus of circumcenter is }\\ &2 x^2+2 y^2-4 x+1=0 \end{aligned} $

Let centre of circle $C$ is $C_1$

Circle $C: x^2+y^2+4 x-6 y-12=0$

With centre $C_2:(-2,3), r=5$

Tangent at $M(1,-1)$ to the circle $C$ is

$ \begin{aligned} & \begin{aligned} & x x_1+y y_1+4\left(\frac{x+x_1}{2}\right) -6\left(\frac{y+y_1}{2}\right)-12=0 \\ & \,\,\,\,\,\,\, \quad x-y+2(x+1)-3(y-1)-12=0 \\ & \Rightarrow \quad 3 x-4 y-7=0 \end{aligned} \end{aligned} $

So, equation of circle $C$ touching the line $3 x-4 y-7=0$ at $(1,-1)$

$ (x-1)^2+(y+1)^2+\lambda(3 x-4 y-7)=0 $

It passes through $(4,0)$

$ \begin{aligned} & (4-1)^2+(0+1)^2+\lambda(12-0-7)=0 \\ & \Rightarrow \quad 9+1+5 \lambda=0 \Rightarrow \lambda=-2 \end{aligned} $

So equation of

$ \begin{gathered} C:(x-1)^2+(y+1)^2-2(3 x-4 y-7)=0 \\ x^2+y^2-8 x+10 y+16=0 \end{gathered} $

Centre $C_1=(4,-5)$

Radius of $C=\sqrt{16+25-16}=5$

Given,

$ C_1: x^2+y^2+2 x+4 y-20=0 $

Centre $O_1(-1,-2), r_1=\sqrt{1+4+20}=5$

$ \begin{aligned} & C_2: x^2+y^2+6 x-8 y+9=0 \\ & \Rightarrow(x+3)^2+(y-4)^2=4^2 \end{aligned} $

Centre $O_2:(-3,4), r_2=\sqrt{9+16-9}=4$

$ r_1+r_2=9 \text { and }\left|r_1-r_2\right|=1 $

Two tangents are there, so $n=2$

$P$ divides $O_1$ and $O_2$ in $5: 4$ ratio externally

$ \begin{gathered} P\left(\frac{(-3 \times 5)-(-1 \times 4)}{5-4}, \frac{(4 \times 5)-(4 \times-2)}{5-4}\right) \\ \equiv P(-11,28) \end{gathered} $

$L$ is the length of tangent from $P$ to $C_2$

$ \begin{aligned} l & =\sqrt{(-11+3)^2+(28-4)^2-4^2} \\ & =\sqrt{64+576-16}=\sqrt{624}=4 \sqrt{39} \\ & \frac{l}{n^2}=\frac{4 \sqrt{39}}{4}=\sqrt{39} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { Let } S_1: x^2+y^2+4 y=0 \\ & C_1(0,-2), r_1=2 \\ & \text { and } S_2: x^2+y^2-4 x-5=0, C_2(2,0) \end{aligned} \end{aligned} $

Common chord of $S_1$ and $S_2$ is

$ \begin{array}{r} S_1-S_2=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\ 4 x+4 y+5=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

$M$ is the point of intersection of common chord and line joining $C_1$ and $C_2$.

Let $M$ divides $C_1 C_2$ in $\lambda$ : 1 ratio internally.

$ M\left(\frac{2 \lambda}{\lambda+1}, \frac{-2}{\lambda+1}\right) $

$M$ lies on common chord

$ \begin{aligned} \frac{8 \lambda}{\lambda+1}-\frac{8}{\lambda+1}+5 & =0 \\ \Rightarrow 8 \lambda-8+5 \lambda+5 & =0 \\ \lambda & =3 / 13 \end{aligned} $

Also $M$ is the centre of the circle whose diameter is common chord of $S_1$ and $S_2$ So, abscissa of

$ M=\frac{2 \lambda}{\lambda+1}=\frac{\frac{6}{13}}{\frac{3}{13}+1}=\frac{6}{16}=\frac{3}{8} $

Let the line cut the axes in $A$ and $B$ and if $(h, k)$ be the mid-point of $A B$, then

$2 h=\frac{1}{\cos \theta}, 2 k=\frac{1}{\sin \theta}$

In order to find the locus, eliminate the variable $\theta$ by $\cos ^2 \theta+\sin ^2 \theta=1$

$\frac{1}{4 h^2}+\frac{1}{4 k^2}=1 \Rightarrow \frac{1}{x^2}+\frac{1}{y^2}=4$

Equation of the given tangents are

$\begin{aligned} E_1=3 x-4 y+4 & =0 \\ E_2=6 x-8 y-7 & =0 \\ \Rightarrow \quad 3 x-4 y-\frac{7}{2} & =0 \end{aligned}$

Here, $a=3, b=-4 c_1=4, c_2=\frac{-7}{2}$

Since, slopes of the given tangents are equal, i.e. $\frac{3}{4}$.

$\therefore$ The given tangents are parallel,

$\begin{aligned} & d=\left|\frac{c_2-c_1}{\sqrt{a^2+b^2}}\right|=\left|\frac{4-\left(\frac{-7}{2}\right)}{\sqrt{3^2+(-4)^2}}\right| \\ & d=\frac{\frac{15}{2}}{\sqrt{9+16}}=\frac{15}{2 \times 5}=\frac{3}{2} \end{aligned}$

As we know that distance between two parallel tangents of a circle is equal to the diameter of that circle.

$\begin{aligned} \therefore \text { Diameter of given circle } & =\frac{3}{2} \\ \therefore \text { Radius of the given circle } & =\frac{\text { Diameter }}{2} \\ & =\left(\frac{3 / 2}{2}\right)=\frac{3}{4} \end{aligned}$

Since, the line $(x-2) \cos \theta+(y-2) \sin \theta=1$ touches a circle. So it is a tangent equation to a circle.

Equation of tangent to a circle at $(x_1, y_1)$ is $(x-h) x_1+(y-k) y_1=a^2$ to a circle

$(x-h)^2+(y-k)^2=a^2$

Then, after comparing

$\begin{aligned} & x-h=x-2 y-k=y-2 \text { and } a^2=1 \\ & x_1=\cos \theta, y_1=\sin \theta \end{aligned}$

$\therefore$ Required equation of circle

$\begin{gathered} (x-2)^2+(y-2)^2=1 \\ \Rightarrow \quad x^2+y^2-4 x-4 y+7=0 \end{gathered}$

Given equation $\left(x^2+y^2-4 x-2 y-20=0\right)$

So, $C=(2,1)$ and radius $=\sqrt{(g)^2+(f)^2-c}$

$=\sqrt{(2)^2+(1)^2+20}$

Radius $=5$

When substituted $x=10$ and $y=7$ in equation, then value becomes

$(10)^2+(7)^2-4(10)-2(7)-20=75$

which is greater than zero.

Thus, the point $(10,7)$ lies outside the circle.

Its distance from the centre of the circle $(2,1)$ is

$\sqrt{(10-2)^2+(7-1)^2}=10 \text { units }$

So, the minimum distance from the circle therefore becomes $10-5=5$ units

Let $x_1$ and $x_2$ are roots of equation $x^2+2 x-a^2=0$ and $y_1$ and $y_2$ are roots of equation $y^2+4 b-b^2=\left(x-x_1\right)\left(x-x_2\right)=0$ and $y^2+4 b-b^2 \rightarrow y_1, y_2=\left(y-y_1\right)\left(y-y_2\right)$ (roots of equation)

Let $A\left(x_1, y_1\right)$ and $B\left(x_2, y_2\right)$.

Now, the equation of circle with diameter $A B$ is given by $\Rightarrow\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0$

$\begin{aligned} & \Rightarrow x^2+2 x-a^2+y^2+4 y-b^2=0 \\ & \Rightarrow x^2+y^2+2 x+4 y=a^2+b^2 \\ & \Rightarrow x^2+y^2+2 x+4 y-a^2-b^2=0 \end{aligned}$

Let $S_1: x^2+y^2-1=0\quad \text{.... (i)}$

$S_2: x^2+y^2-8 x+15=0\quad \text{.... (ii)}$

and $S_3: x^2+y^2+10 y+24=0\quad \text{.... (iii)}$

From Eqs. (i) and (ii),

$\begin{aligned} & 8 x-15=1 \\ \Rightarrow \quad & x=2 \end{aligned}$

Now, from Eqs. (i) and (iii),

$\begin{aligned} & -10 y-24=1 \\ & \Rightarrow \quad-10 y=25 \Rightarrow y=\frac{-5}{2} \end{aligned}$

Hence, the radical centre is $\left(2, \frac{-5}{2}\right)$.

$\begin{aligned} \text { Let } x & =\frac{8 t}{1+t^2} \quad \text{.... (i)}\\ y & =\frac{4\left(1-t^2\right)}{1+t^2} \quad \text{.... (ii)} \end{aligned}$

Squaring and adding Eqs. (i) and (ii), we get

$\begin{aligned} x^2+y^2 & =\frac{64 t^2}{\left(1+t^2\right)^2}+\frac{16\left(1-2 t^2+t^4\right)}{\left(1+t^2\right)^2} \\ & =\frac{16\left(1+2 t^2+t^4\right)}{\left(1+t^2\right)^2}=\frac{16\left(1+t^2\right)^2}{\left(1+t^2\right)^2}=16 \\ \Rightarrow x^2+y^2 & =(4)^2 \end{aligned}$

which is circle of radius 4 units.

Centre of circle passing through $(1,2),(5,2)$ and $(5,-2)$ is mid-point of $C(1,2)$ and $A(5,-2)$. We know that diameter always form a right angle on a circle.

$\therefore \text { Centre }=\left(\frac{1+5}{2}, \frac{2-2}{2}\right)=(3,0)$

$\begin{aligned} \text { and Radius } & =\frac{\text { Diameter }}{2} \\ & =\frac{\sqrt{(5-1)^2+(-2-2)^2}}{2}=2 \sqrt{2} \end{aligned}$

Thus, area is $\pi(2 \sqrt{2})^2=8 \pi$ (units)$^2$

To find the shortest and longest distances from the point $(2,-7)$ to the circle

$x^2+y^2-14 x-10 y-151=0$

So, to find the solution we first draw the diagram,

So, we have to find the coordinates of the centre 0 and the radius of the circle, i.e $O M=O N=$ radius Thus, the shortest distance $P M=O M-O P$ and the longest distance $P N=M N-P M$ First we have to determine the point $P(2,-7)$ lies in which side of circle. So, we put the value $x=2,y=-7$ in the left hand side of the equation, we get $x^2+y^2-14 x-10 y-151 =(2)^2$

$=(2)^2+(-7)^2-14\times2-10(-7)-151=-56 < 0$

Thus, the point lies inside of the circle. Thus, we know any equation of the circle is in the from $x^2+y^2+2 g x+2 f y+c=0$.

Thus, $g=-7$, and $f=-5, c=-151$

So, centre is $(-g,-f)=(7,5)$

$\begin{aligned} \text { and radius } & =r=O M=O N \\ & =\sqrt{7^2+5^2-(-151)} \\ & =\sqrt{49+25+151} \\ & =\sqrt{225}=15 \end{aligned}$

Now, distance between centre $O(7,5)$ and $P(2,-7)$ is $O P=\sqrt{(7-2)^2+(5+7)^2}$

$=\sqrt{25+144}=\sqrt{169}=13$

Thus, the shortest distance from the point $P(2,-7)$ to the circle is $P M=O M-O P=15-13=2$ units and the longest distance is

$\begin{aligned} P N=M N-P M & =(O M+O N)-P M \\ & =(15+15)-2=28 \text { units } \end{aligned}$

Thus, the required ratio is $28: 2=14: 1$

Let the centre of circle be $(h, k)$. Equation of circle will be

$x^2+y^2-2 h x-2 k y+c=0 \quad \text{... (i)}$

Circle passes through $(2,3)$

$\begin{array}{llrl} & & 2^2+3^2-4 h-6 k+c & =0 \\ \Rightarrow & & 4 h+6 k-c=13 \\ \text { or } & & c=4 h+6 k-13 \end{array}$

Equation of circle be

$x^2+y^2-2 h x-2 k y+4 h+6 k-13$

This circle makes intercept of length 3 on the line $x=2$

$\left|y_1-y_2\right|=3$, when $x=2$ is substituted in the equation of circle, where $y_1$ and $y_2$ are the two roots

$\begin{aligned} 4+ & y^2-4 h-2 k y+4 h+6 k-13=0 \\ \Rightarrow & y^2-2 k y+6 k-9=0 \\ & y_1+y_2=2 k \text { and } y_1 y_2=6 k-9 \\ & \left|y_1-y_2\right|=3 \\ & \sqrt{\left(y_1+y_2\right)^2-4 y_1 y_2}=3 \\ \Rightarrow & 4 k^2-4(6 k-9)=9 \\ \Rightarrow & 4 k^2-24 k+36-9=0 \Rightarrow 4 k^2-24 k+27=0 \\ \Rightarrow & k=\frac{9}{2}, k \neq \frac{3}{2} \end{aligned}$

[$\because$ when substitute $k=\frac{3}{2}$ in $y_1 y_2$, we get $y_1 y_2$ as 0$]$

Now, similarly substitute $y=3$

$\begin{aligned} & x^2+9-2 h x-6 k+4 h+6 k-13=0 \\ \Rightarrow & x^2-2 h x+4 h-4=0 \\ \Rightarrow & x_1+x_2=2 h_1, x_1 x_2=4 h-4 \\ \because \quad & \left|x_1-x_2\right|=4 \\ & \sqrt{\left(x_1+x_2\right)^2-4 x_1 x_2}=4 \\ \Rightarrow \quad & 4 h^2-16 h+16=16 \Rightarrow 4 h(h-4)=0 \\ & h=4 \text { and } h \neq 0 \end{aligned}$

$\therefore$ Equation of the required circle is

$x^2+y^2-8 x-9 y+30=0$

The given equation of circles

$\begin{aligned} & S_1: x^2+y^2+8 x+2 y+10=0 \\ & S_2: x^2+y^2+7 x+3 y+10=0 \end{aligned}$

As we know the equation of radical axis of two circles is given by

$\begin{aligned} S_1- & S_2=0 \\ \Rightarrow & \left(x^2+y^2+8 x+2 y+10\right) \\ & -\left(x^2+y^2+7 x+3 y+10\right)=0 \\ \Rightarrow & (8 x-7 x)+(2 y-3 y)=0 \\ \Rightarrow & x-y=0 \end{aligned}$

Thus, equation of radical axis of the given circle is

$x-y=0$

Hence, the image of the point $(3,4)$ with respect to $y=x$ line is $(4,3)$.

Common tangents which are parallel are

$\begin{array}{ll} & 3 x-4 y+4=0 \\ \text { and } & 6 x-8 y-7=0 \\ \Rightarrow & 3 x-4 y+4=0 \\ \text { and } & 3 x-4 y-\frac{7}{2}=0 \end{array}$

The diameter is equal to the distance between these parallel lines.

So, the radius $=\frac{1}{2} \times\left|\frac{4-\left(-\frac{7}{2}\right)}{\sqrt{3^2+(-4)^2}}\right|$

$=\frac{1}{2} \times \frac{15}{2} \times \frac{1}{5}=\frac{3}{4}$

The centre of the circle lies on the line parallel to the given lines at a distance of $\frac{3}{4}$ from both the lines.

So, let the equation be $3 x-4 y+k=0\quad \text{....(i)}$

Then, distance between $3 x-4 y+k=0$ and $3 x-4 y+4$ is same as distance between $3 x-4 y+k=0$ and $3 x-4 y-\frac{7}{2}=0$

$\begin{aligned} & \left|\frac{k-4}{\sqrt{3^2+(-4)^2}}\right|=\left|\frac{k-\left(-\frac{7}{2}\right)}{\sqrt{(3)^2+(-4)^2}}\right| \\ & \Rightarrow \quad|k-4|=-\left|k+\frac{7}{2}\right| \\ & \text { and }-(k-4)=k+\frac{7}{2} \\ & \therefore \quad 2 k=\frac{7}{2}-4=-\frac{1}{2} \Rightarrow k=-\frac{1}{4} \\ & \text { and } 2 k=4-\frac{7}{2}=\frac{1}{2} \Rightarrow k=\frac{1}{4} \end{aligned}$

$\therefore$ For $k=\frac{1}{4}$ distance of Eq. (i) from other line is $\frac{3}{4}$.

$\therefore$ Locus of centers of the circles, is

$3 x-4 y+\frac{1}{4}=0$

Curve : $x^2+y^2=1$

Let $\left(\frac{1}{a}, \frac{1}{b}\right)=\left(x_1, y_1\right)$ touches the circles, then equation of tangent at $\left(x_1, y_1\right)$ is

$\begin{aligned} x x_1+y y_1 & =1 \\ \Rightarrow \quad x \cdot\left(\frac{1}{a}\right)+y \cdot\left(\frac{1}{b}\right) & =1 \Rightarrow \frac{x}{a}+\frac{y}{b}=1 \end{aligned}$

$\begin{aligned} & 4 x-3 y-10=0 \\ & \Rightarrow \quad x=\frac{3 y+10}{4} \\ & \text { and } x^2+y^2-2 x+4 y-20=0 \\ & \Rightarrow \quad\left(\frac{3 y+10}{4}\right)^2+y^2-2\left(\frac{3 y+10}{4}\right)+4 y-20=0 \\ & \Rightarrow \quad 9 y^2+100+60 y+16 y^2-24 y-80 +64y-320=0\\ & \Rightarrow \quad 25 y^2+100 y-300=0 \Rightarrow y^2+4 y-12=0 \\ & \Rightarrow \quad y^2+6 y-2 y-12=0 \Rightarrow y(y+6)-2(y+6)=0 \\ & \Rightarrow \quad(y+6)(y-2)=0 \\ & y=-6,2 \\ & y=-6 \Rightarrow x=\frac{-18+10}{4}=-2 \Rightarrow(-2,-6) \\ & y=2 \Rightarrow x=\frac{6+10}{4}=4 \Rightarrow(4,2) \end{aligned}$

Required length

$\begin{aligned} & =\sqrt{(4+2)^2+(2+6)^2} \\ & =\sqrt{36+64}=\sqrt{100}=10 \end{aligned}$

Let the pole be $\left(x_1, y_1\right)$, then the equation of the polar with respect to the circle $x^2+y^2=c^2$ is $x x_1+y y_1=c^2$

$\Rightarrow \quad \frac{x x_1}{c^2}+\frac{y y_1}{c^2}=1 \quad \text{... (i)}$

Eq. (i) and $\frac{x}{a}+\frac{y}{b}=1$ represents the same line.

$\begin{aligned} \therefore \quad \frac{x_1}{c^2} & =\frac{1}{a} \Rightarrow x_1=\frac{c^2}{a} \\ \frac{y_1}{c^2} & =\frac{1}{b} \Rightarrow y_1=\frac{c^2}{b} \end{aligned}$

Pole : $\left(x_1, y_1\right)=\left(\frac{c^2}{a}, \frac{c^2}{b}\right)$

Line: $5 x-2 y+6=0$

On $Y$-axis, $x=0 \Rightarrow y=\frac{6}{2}=3$

i.e. $Q \equiv(0,3)$

$\begin{aligned} & \text { Circle : } x^2+y^2+6 x+6 y-2=0 \\ & \text { Centre }(-g,-f) \equiv(-3,-3) \\ & \text { Radius }=\sqrt{g^2+f^2-c} \\ & =\sqrt{9+9+2}=\sqrt{20} \\ & =2 \sqrt{5} \\ & C Q=\sqrt{(0+3)^2+(3+3)^2}=\sqrt{9+36}=\sqrt{45} \\ & \text { In } \triangle P Q C \text {, } \\ & (C Q)^2=(C P)^2+(Q P)^2 \\ & \Rightarrow \quad 45=20+(Q P)^2 \\ & \Rightarrow \quad 25=(Q P)^2 \Rightarrow Q P=5 \\ & \end{aligned}$

Let $P\left( {t,{{4t - 20} \over 5}} \right)$ be a point on the line $4x - 5y = 20$. Then the chord of contact of tangents drawn from P to the circle ${x^2} + {y^2} = 9$ is

$tx + \left( {{{4t - 20} \over 5}} \right)y = 9$ ...... (i)

Let $\theta$ (h, k) be the mid-point of this chord of contact, then its equation is also $hx + ky = {h^2} + {k^2}$ ..... (ii) [using $T = S'$]

Clearly Eqs. (i) and (ii) represent the same line

$\therefore$ ${t \over h} = {{4t - 20} \over {5k}} = {9 \over {{h^2} + {k^2}}}$

$ \Rightarrow {t \over h} = {9 \over {{h^2} + {k^2}}}$ and ${t \over h} = {{4t - 20} \over {5k}}$

$ \Rightarrow t = {{9h} \over {{h^2} + {k^2}}}$ and $t = {{20h} \over {4h - 5k}}$

$ \Rightarrow {{9h} \over {{h^2} + {k^2}}} = {{20h} \over {4h - 5k}}$

$ \Rightarrow h\{ 20({h^2} + {k^2}) - 36h + 45k\} = 0$

$x = 0$ or $[20({x^2} + {y^2}) - 36x + 45y] = 0$