Question

Tangent OA and OB are drawn for $O(0,0)$ to the circle $(x-1{)}^2+(y-1{)}^2=1$.
Equation of the circumcircle of triangle OAB is

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

Answer 1

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

Circumcircle of the triangle OAB is the circle with A and B as the ends of diameter.
let the point of contact of the tangent from $O(0,0)$ be $G(h,k)$.
$\Delta OGC$ is right angled. Hence by Pythagoras theorem $h^2+k^2=1$
The distance CG is radius where C is the center of the circle.
$\Rightarrow (h-1{)}^2+(k-1{)}^2=1$
$\Rightarrow h+k=1$
So, $G=(1,0)$ or $(0,1)$
Hence $A(1,0)B(0,1)$
Equation of circle with A and B as the ends of diameter is $(x-1)\left(x\right)+\left(y\right)(y-1)=0.$