Question

If a circle passes through the point $ (1,\,2) $ and cuts the circle $ x^{2}+y^{2}=4 $ orthogonally, then the equation of the locus of its centre is

• A. $x^{2}+y^{2}-3\,x-8\,y+1=0$
• B. $x^{2}+y^{2}-2\,x-6\,y-7=0 $
• C. $ 2\,x+4\,y-9=0 $
• D. $ 2\,x+4\,y-1=0 $

Answer 1

Answer: (C)

Let the equation of the circle be
$x^{2}+y^{2}+2 \,g x+2 \,f y+ c=0$
Since, this passes through $(1,2)$.
$\therefore 1^{2}+2^{2}+2\, g(1)+2\, f(2)+c=0$
$\Rightarrow 5+2\, g+4 \,f+ c=0 \ldots(i)$
Also, the circle $x^{2}+y^{2}=4$ intersects the circle
$x^{2}+y^{2}+2 \,g x+2\, f y +c=0$ orthogonally.
$\therefore 2(g \cdot 0+f \cdot 0)=c-4$
$\Rightarrow c=4$
On putting the value of $c$ in Eq. (i), we get
$2 \,g+4 \,f+9=0$
Hence, the locus of centre $(-g,-f)$ is
$-2\,x-4\,y+9=0$
or $2\,x+4\,y-9=0$