Question

The circle $C_1:x^2+y^2=8$ cuts orthogonally the circle $C_2$ whose centre lies on the line $x-y-4=0$ then, the circle $C_2$ passes through a fixed point, which lies on

• A. $x-2y=0$
• B. $x+y=0$
• C. $x-2y=0$
• D. $x+2y=0$

Answer 1

The correct option is B $x+y=0$

The centre of $C_2$ be $(g,f)$.
$g-f-4=0\Rightarrow g=f+4$
Equation of $C_2:x^2+y^2-2gx-2fy+k=0$
$C_2:x^2+y^2-2(f+4)x-2fy+k=0$
$C_1:x^2+y^2-8=0$ are orthogonal.
Orthogonality condition -
$2\times 0\times (f+4)+2\times 0\times \left(f\right)=k-8$
$\Rightarrow k=8$
$C_2:x^2+y^2-2fx-8x-2fy+8=0\Rightarrow x^2+y^2-8x+8-2f(x+y)=0$
Combination of circle $x^2+y^2-8x+8=0$ and line $x+y=0$.
$\Rightarrow C_2$ passes through the point of intersection of the circle and the straight line $x+y=0$.

$y=-x\&x^2+y^2-8x+8-2f(x+y)=0\Rightarrow x^2+x^2-8x+8=0\Rightarrow (x-2{)}^2=0$
$\Rightarrow x=2;y=-2$ is the point of intersection.