Two circles {x^2} + {y^2} - 2x + 22y + 5 = 0 and {x^2} + {y^2} + 14x + 6y + k = 0 intersect orthogon

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10-1.Circle and System of Circles

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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

A

$47$

B

$ - 47$

C

$49$

D

$ - 49$

Solution

(a) Given two circles

${x^2} + {y^2} – 2x + 22y + 5 = 0$

${x^2} + {y^2} + 14x + 6y + k = 0$

The two circles cut orthogonally, if

$2({g_1}{g_2} + {f_1}{f_2}) = {c_1} + {c_2}$

$i.e.$, $2( – 1.7 + 11.3) = 5 + k$

$2( – 7 + 33) = 5 + k $

$\Rightarrow 52 – 5 = k$

$\Rightarrow k = 47$.

Standard 11

Mathematics

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