Two circles $x^{2} + y^{2} + p x + p y - 7 = 0$ and $x^{2} + y^{2} - 10 x + 2 p y + 1 = 0$ will cut orthogonally if the value of $p$ is

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Solution

The correct options are
C $2$
D $3$
We know that two circles will cut orthogonally if $2 f_{1} f_{2} + 2 g_{1} g_{2} + c_{1} + c_{2} = 0$
$\Rightarrow 2 (\frac{p}{2}) (\frac{- 10}{2}) + 2 (\frac{p}{2}) (\frac{2 p}{2}) - 7 + 1 = 0$
$\Rightarrow p^{2} - 5 p + 6 = 0$
$∴ p = 2$ or $3$

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