Question

Tangents $OA$ and $OB$ are drawn to the circle $x^2+y^2+gx+fy+c=0$ from $O(0,0)$ . The equation of the circumcircle of the $\Delta$ $OAB$ is

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

Answer 1

The correct option is D $2x^2+2y^2+gx+fy=0$

Angle made by semicircle is right angle.

$\therefore$ point $O(0,0)$ and $C(-\frac{g}{2},-\frac{f}{2})$

are diameter ends of a circle passing through $O,B\&A$.

So, equation of circumcircle of the $\Delta OAB$ is

$(x-0)(x+\frac{g}{2})+(y-0)(y+\frac{f}{2})=0$
$\Rightarrow x^2+y^2+\frac{xg}{2}+\frac{yf}{2}=0$

$\Rightarrow 2x^2+2y^2+xg+yf=0$