Question

The centre of the circle $S$ lies on $2x-2y+9=0$ and it cuts orthogonally the circle $x^2+y^2=4$. then the circle passes through two fixed points

• A. $(1,1),(3,3)$
• B. $(-\frac{1}{2},\frac{1}{2})(-4,4)$
• C. $(0,0),(5,5)$
• D. None of these

Answer 1

Answer: (B)

$(-\frac{1}{2},\frac{1}{2})(-4,4)$

Let $S=x^2+y^2+2gx+2fy+c=0$
$\because$ it cuts $x^2+y^2=4$ orthogonally
$\Rightarrow c=4$
Moreover $-2g+2f+g=0$
$(\because (-g,-f)$ satisfy the given equation $)$
$\therefore S\equiv x^2+y^2+2gx+2fy+4=0\Rightarrow x^2+y^2+(2f+9)x+2fy+4=0\Rightarrow (x^2+y^2+9x+4)+2f(x+y)=0$
It is of the form $S+\lambda P=0$ and hence passes through the intersection of $S=0$ and $P=0$ which when solved give $(-\frac{1}{2},\frac{1}{2})(-4,4)$