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(C) The given equation of the circle is $x^2 + 2px + y^2 - 2qy + q^2 = 0$.Rewriting this,we get $(x + p)^2 - p^2 + (y - q)^2 - q^2 + q^2 = 0$,which si

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$p^2 - q^2 = 0$

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(C) The given equation of the circle is $x^2 + 2px + y^2 - 2qy + q^2 = 0$.Rewriting this,we get $(x + p)^2 - p^2 + (y - q)^2 - q^2 + q^2 = 0$,which simplifies to $(x + p)^2 + (y - q)^2 = p^2$.The center of the circle is $(-p, q)$ and the radius is $|p|$.Since the circle passes through the origin $(0, 0)$ (as $0^2 + 2p(0) + 0^2 - 2q(0) + q^2 = q^2 
eq 0$,wait,let's re-evaluate).Actually,the equation $x^2 + 2px + y^2 - 2qy + q^2 = 0$ at $(0,0)$ gives $q^2 = 0$,so $q=0$. If $q=0$,the circle is $x^2 + 2px + y^2 = 0$,which passes through the origin.For the tangents from the origin to be perpendicular,the origin must lie on the circle,and the radius must be such that the angle between tangents is $90^\circ$.Alternatively,the condition for perpendicular tangents from the origin to a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $g^2 + f^2 = 2c$.Here,$2g = 2p \Rightarrow g = p$,$2f = -2q \Rightarrow f = -q$,and $c = q^2$.Substituting these into the condition: $p^2 + (-q)^2 = 2(q^2)$ $\Rightarrow p^2 + q^2 = 2q^2$ $\Rightarrow p^2 = q^2$ $\Rightarrow p^2 - q^2 = 0$.

When are the tangents drawn from the origin to the circle $x^2 + 2px + y^2 - 2qy + q^2 = 0$ perpendicular to each other?

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