Question

If $O$ is the origin and $OP,OQ$ are the tangents from the origin to the circle $x^2+y^2-6x+4y+9=0$, then circumcenter of the triangle $OPQ$ is

• A. $(3,-2)$
• B. $(\frac{3}{2},-1)$
• C. $(\frac{3}{4},\frac{1}{2})$
• D. $(-\frac{3}{2},1)$

Answer 1

The correct option is B $(\frac{3}{2},-1)$

Clearly, OPCQ is cyclic quadrilateral, then circumcircle of $\Delta$OPQ passes through the point C.
Since OP and OQ are tangents to the circle they make perpendicular with CP and CQ respectively. Also a circle inscribes two right angled triangles which are COP and COQ in this case.
Thus for this circle, OC is a diameter, then centre is midpoint of OC which is $(\frac{3}{2},-1)$.