Tangents OP and OQ are drawn from the origin | vedclass

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(B) The equation of the circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. Since the tangents are drawn from the origin $(0,0)$,the equation of the chord of c

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$x^2 + y^2 + gx + fy = 0$

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(B) The equation of the circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. Since the tangents are drawn from the origin $(0,0)$,the equation of the chord of contact $PQ$ is given by $T = 0$,which is $gx + fy + c = 0$. Any circle passing through the intersection of the circle and the chord of contact is given by $(x^2 + y^2 + 2gx + 2fy + c) + \lambda(gx + fy + c) = 0$. Since this circle passes through the origin $(0,0)$,we substitute $x=0, y=0$: $(0^2 + 0^2 + 0 + 0 + c) + \lambda(0 + 0 + c) = 0 $ $\Rightarrow c + \lambda c = 0 $ $\Rightarrow \lambda = -1$ (assuming $c 
eq 0$). Substituting $\lambda = -1$ into the equation: $(x^2 + y^2 + 2gx + 2fy + c) - (gx + fy + c) = 0 \Rightarrow x^2 + y^2 + gx + fy = 0$.

Tangents $OP$ and $OQ$ are drawn from the origin $O$ to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$. Then,the equation of the circumcircle of the triangle $OPQ$ is

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