Question

The radius of the circle which touches the line $x+y=0$ at $M(-1,1)$ and cuts the circle $x^2+y^2+6x-4y+18=0$ orthogonally, is

• A. $3\sqrt{2}$
• B. $4\sqrt{2}$
• C. $\sqrt{2}$
• D. $5\sqrt{2}$

Answer 1

The correct option is D $5\sqrt{2}$

Clearly, equation of the required circle is
$(x+1{)}^2+(y-1{)}^2+\lambda (x+y)=0\Rightarrow x^2+y^2+(\lambda +2)x+(\lambda -2)y+2=0\dots \left(1\right)$


As circle $\left(1\right)$ intersects the circle $x^2+y^2+6x-4y+18=0$ orthogonally, so using orthogonality condition $2(g_1{g}_2+f_1{f}_2)=c_1+c_2$, we get
$2[\left(\frac{\lambda +2}{2}\right)3-2\left(\frac{\lambda -2}{2}\right)]=18+2$
$\Rightarrow (3\lambda +6)-(2\lambda -4)=20\Rightarrow \lambda =10$
Putting $\lambda =10$ in equation $\left(1\right)$, we get
$x^2+y^2+12x+8y+2=0.$
Radius $=\sqrt{6^2+4^2-2}$
$=\sqrt{50}=5\sqrt{2}$