Question

Find k is the following pair of circles are orthogonal x2+y2-6x-8y+12=0,x2+y2-16y+k=0
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Answer 1

EXPLANATION.

pair of circle are orthogonal

=> x² + y² - 6x - 8y + 12 = 0

=> x² + y² - 16y + k = 0

$\sf : \implies \: two \: circle \: are \: \: orthogonal \: if \\ \\ \sf : \implies \: 2 g_{1}g_{2} \: + \: 2 f_{1}f_{2} \: = c_{1} \: + \: c_{2}$

$\sf : \implies \: {x}^{2} + {y}^{2} - 6x - 8y + 12 = 0 \\ \\ \sf : \implies \: g_{1} \: = - 3 \: \:and \: \: \: \: f_{1} \: = - 4 \: \: \: and \: \: \: c_{1} = 12$

$\sf : \implies \: {x}^{2} + {y}^{2} - 16y + k = 0 \\ \\ \sf : \implies \: g_{2} = 0 \: \: \: \: and \: \: \: f_{2} = - 8 \: \: \: \: and \: \: \: \: c_{2} = k$

$\sf : \implies \: 2( - 3)( 0) + 2( - 4)( - 8) = 12 + k \\ \\ \sf : \implies \: 0 + 64 = 12 + k \\ \\ \sf : \implies \: 64 - 12 = k \\ \\ \sf : \implies \: 52 = k \:$

$\sf : \implies \: \green{{ \underline{value \: of \: k \: = 52}}}$