Question

The circle $x^2+y^2=4$ cuts the circle $x^2+y^2+2x+3y-5=0$ in $A\&B$ . Then the equation of the circle on $AB$ as a diameter is :-

• A. $13(x^2+y^2)-4x-6y-50=0$
• B. $9(x^2+y^2)+8x-4y+25=0$
• C. $x^2+y^2-5x+2y+72=0$
• D. $13(x^2+y^2)-4x-6y+50=0$

Answer 1

Answer: (A)

$13(x^2+y^2)-4x-6y-50=0$

Given:$S_1:x^2+y^2-4=0$ and $S_2:x^2+y^2+2x+3y-5=0$ cuts the circle at the points $A$ and $B$

Join $AB$

$\Rightarrow AB$ is a common chord to the circle.

$\Rightarrow$Common chord equation is

$S_1-S_2=0$

$\Rightarrow x^2+y^2-4-(x^2+y^2+2x+3y-5)=0$

$\Rightarrow x^2+y^2-4-x^2-y^2-2x-3y+5=0$

$\Rightarrow -2x-3y+1=0$

$\Rightarrow 2x+3y-1=0$ is the equation of the line $AB$

Equation of diameter of the circle is $S_1+\lambda L=0$

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

$\Rightarrow x^2+y^2+2x\lambda +3y\lambda -\lambda -4=0$ .........$\left(1\right)$

$\Rightarrow$Centre of the circle is $(-\lambda ,\frac{-3\lambda }{2})$ lies on the line $AB$

$\Rightarrow (-\lambda ,\frac{-3\lambda }{2})$ lies on the line $2x+3y-1=0$

$\Rightarrow 2\times -\lambda +3\times \frac{-3\lambda }{2}-1=0$

$\Rightarrow -2\lambda -\frac{9\lambda }{2}=1$

$\Rightarrow \frac{-4\lambda -9\lambda }{2}=1$

$\Rightarrow \frac{-13\lambda }{2}=1$

$\therefore \lambda =\frac{-2}{13}$

Substituting $\lambda =\frac{-2}{13}$ in equation $\left(1\right)$ we get

$\Rightarrow x^2+y^2+2x\times \frac{-2}{13}+3y\times \frac{-2}{13}-\left(\frac{-2}{13}\right)-4=0$

$\Rightarrow x^2+y^2-\frac{4}{13}x-\frac{6}{13}y+\frac{2}{13}-4=0$

$\Rightarrow x^2+y^2-\frac{4}{13}x-\frac{6}{13}y+\frac{2-52}{13}=0$

$\Rightarrow x^2+y^2-\frac{4}{13}x-\frac{6}{13}y-\frac{50}{13}=0$

$\therefore 13(x^2+y^2)-4x-6y-50=0$ is the required of the circle on $AB$ as diameter.