Question

Find the equation of the circle passing through the points of intersection of the circles $x^2+y^2-2x-4y-4=0$ and $x^2+y^2-10x-12y+40=0$ and whose radius is 4

Answer 1

$\to$ Circle would be of form $C_1+\lambda C_2=0$

$\Rightarrow (1+\lambda )x^2+(1+\lambda )y^2+(-2-10\lambda )x+(-4-12\lambda )y-4+40\lambda =0$

Radius $=\sqrt{(\frac{-2-10\lambda }{2}{)}^2+(\frac{-4-12\lambda }{2}{)}^2+4-40\lambda }$

$\Rightarrow \sqrt{(1+5\lambda {)}^2+(2+6\lambda {)}^2+4-40\lambda =4}$

$\Rightarrow \sqrt{25{\lambda }^2+36{\lambda }^2-40\lambda +10\lambda +24\lambda +9}=4$

$\Rightarrow 61{\lambda }^2-6\lambda +9=16$

$\Rightarrow 61{\lambda }^2-6\lambda -7=0$

Get $\lambda =\frac{6\pm \sqrt{36+28\left(61\right)}}{2\times 61}$ & circles equation.