Question

Tangents are drawn from the point $P(2,2)$ to the circle $x^2+y^2=1,$ touching the circle at $A$ and $B.$ Then equation of circumcircle of $△PAB$ is

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

Answer 1

The correct option is B $x^2+y^2-2x-2y=0$


For the required circle, $P(2,2)$ and $(0,0)$ will be end points of its diameter.
Hence, equation will be
$(x-2)(x-0)+(y-0)(y-2)=0\Rightarrow x^2+y^2-2x-2y=0$