The two circles $x^{2} + y^{2} - 2x + 22y + 5 = 0$ and $x^{2} + y^{2} + 14x + 6y + k = 0$ intersect orthogonally provided $k$ is equal to

Question

Dr. Abubacker · Accepted Answer

(a)
Here, $g_{1} = - 1,\ f_{1} = 11,\ c_{1} = 5$
and $g_{2} = 7,\ f_{2} = 3,\ c_{2} = k$
$\Rightarrow \ \ \ 2( - 1.7 + 11.3) = 5 + k\ \  \Rightarrow k = 47$

Q. The two circles $x^{2} + y^{2} - 2x + 22y + 5 = 0$ and $x^{2} + y^{2} + 14x + 6y + k = 0$ intersect orthogonally provided $k$ is equal to

Explanation

Reviewed by Dr. Abubacker

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