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

Answer 1

Answer: (C)

Let the equation of the circle be
$x^2+y^2+2gx+2fy+c=0$
Since, this passes through $(1,2)$.
$\therefore 1^2+2^2+2g(1)+2f(2)+c=0$
$\Rightarrow 5+2g+4f+c=0\dots (i)$
Also, the circle $x^2+y^2=4$ intersects the circle
$x^2+y^2+2gx+2fy+c=0$ 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
$2g+4f+9=0$
Hence, the locus of centre $(-g,-f)$ is
$-2x-4y+9=0$
or $2x+4y-9=0$