Question

The line $x+2y+a=0$ intersects the circle $x^2+y^2-4=0$ at two distinct points $A$ and $B$. Another line $12x-6y-41=0$ intersects the circle $x^2+y^2-4x-2y+1=0$ at two distinct points $C$ and $D$.
The equation of the circle passing through the points $A,B,C$ and $D$ is

• A. $5x^2+5y^2-8x-16y-36=0$
• B. $5x^2+5y^2+8x-16y-36=0$
• C. $5x^2+5y^2=8x+16y-36=0$
• D. $5x^2+5y^2-8x-16y+36=0$

Answer 1

The correct option is A $5x^2+5y^2-8x-16y-36=0$

$L_1=x^2+y^2-4=0$

$L_2=x^2+y^2-4x-2y+1=0$

Radical axis of $C_1$ and $C_2$ is $:$

$4x+2y=5$-------------$\left(i\right)$

$4x+2y-5=0$-----------$\left(ii\right)$

$x+2y+a=0$-------------$\left(iii\right)$

If $P,Q,R,S$ are concylic then $\left(i\right),\left(ii\right),\left(iii\right)$ are concurvet

$\left|\begin{array}{c}42-5\\ 12a\\ 12-6-41\end{array}\right|=0$

$4\left(-82+6a\right)-2\left(-41-12a\right)-5\left(-6-24\right)=0$

$48a=82\times 3-150$

$48a=96$

$a=2$

The equation of finally of circle passing them $PQRS$ is

$L_1+\lambda L_1=0$

$x^2+y^2-4+\lambda \left(x+2y+2\right)=0$-------------$\left(IV\right)$

$L_2+\lambda L_2=0$

$x^2+y^2-4x-2y+1+\mu \left(12x-6y-41\right)=0$------------$\left(V\right)$

on comparing $\left(IV\right)$ and $\left(V\right)$

$\lambda =12\mu -4$

$2\lambda -4=1-41\mu$

$65\mu =13$

$\mu =\frac{1}{5}$

$\lambda =\frac{12}{5}-4$

$\lambda =\frac{-8}{5}$

Hence equation of circle is

$x^2+y^2-4\times \frac{-8}{5}\left(x+2y+2\right)=0$

$5x^2+5y^2-8x-16y-36=0$

Hence$,$ option $\left(A\right)$ is correct$.$