The tangents drawn from the origin to the circle $x^{2} + y^{2} - 2 p x - 2 q y + q^{2} = 0$ are perpendicular if

p = q

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p^{2} = q^{2}

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q = - p

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p^{2} + q^{2} = 1

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Solution

The correct options are
A $p = q$
B $p^{2} = q^{2}$
C $q = - p$
Equation of the given circle can be written as $(x - p)^{2} + (y - q)^{2} = p^{2}$
so, that the centre of the circle is $(p, q)$ and its radius is $p$ .
This shows that $x = 0$ is a tangent to the circle from the origin.
Since tangents from the origin are perpendicular, the equation of the other tangent must be $y = 0$ ,
which is possible if $q = \pm p$ or $p^{2} = q^{2}$

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