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

h = r

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h = - r

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r^{2} + h^{2} = 1

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r^{2} + h^{2} = 2

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Solution

The correct options are
B $h = - r$
C $h = r$
Equation of the given circle can be written as ${(x - r)}^{2} + {(y - h)}^{2} = p^{2}$
This has $(r, h)$ as the center and $r$ as the radius showing that it touches $y -$ axis.
$\Rightarrow$ Other tangent from the origin to the circle must be $x -$ axis which is possible if $h = \pm r$

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