If O is the origin and OP,OQ are tangents to | vedclass

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(D) Let the circle be $S \equiv x^2 + y^2 + 2gx + 2fy + c = 0$. The origin is $O(0, 0)$.The tangents $OP$ and $OQ$ are drawn from the origin to the ci

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None of these

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(D) Let the circle be $S \equiv x^2 + y^2 + 2gx + 2fy + c = 0$. The origin is $O(0, 0)$.The tangents $OP$ and $OQ$ are drawn from the origin to the circle.The points of contact $P$ and $Q$ lie on the circle,and the line $PQ$ is the chord of contact.The equation of the chord of contact $PQ$ is $T = 0$,which is $gx + fy + c = 0$.The circle passing through the points of intersection of the circle $S=0$ and the line $PQ=0$ is given by $S + \lambda(gx + fy + c) = 0$.Since the circle passes through the origin $(0, 0)$,we have $c + \lambda(c) = 0$,which implies $\lambda = -1$.Substituting $\lambda = -1$,the equation of the circle passing through $O, P, Q$ is $x^2 + y^2 + 2gx + 2fy + c - (gx + fy + c) = 0$,which simplifies to $x^2 + y^2 + gx + fy = 0$.The circumcentre of $	riangle OPQ$ is the centre of this circle,which is $\left(-\frac{g}{2}, -\frac{f}{2}ight)$.Since this value is not among the given options,the correct answer is $(d)$.

If $O$ is the origin and $OP$,$OQ$ are tangents to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$,the circumcentre of the triangle $OPQ$ is

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