Question

If O is the origin and OP, OQ are the tangents from the origin to the circle $x^2+y^2-6x+4y+8=0$, the 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 $(\frac{3}{2},-1)$

Now PQ is the chord of contact of the tangents from the origin to the circle $x^2+y^2-6x+4y+8=0\to \left(1\right)$
Equation of PQ is $3x-2y-8=0\to \left(2\right)$
Equation of a circle passing through the intersection of (1) and (2) is
$x^2+y^2-6x+4y+8+\lambda (3x-2y-8)=0\to \left(3\right)$
if this represents the circumcircle of the triangle OAB, it passes through O(0, 0), so $\lambda =1$ and the equation (3) becomes $x^2+y^2-3x+2y=0$
$\therefore$ The required coordinates of the centre are $(\frac{3}{2},-1)$