Question

Find the equation of the circle with centre : (-a, -b) and radius √a2−b2​

Answer 1

Answer:

h = t + 9-5(7-9) = 8

Step-by-step explanation:

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Answer 2

Answer:

hey here is your answer

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Step-by-step explanation:

$so \: for \: a \: certain \: circle \\ coordinates \: of \: its \: centre \: are \: ( - a, - b) \\ so \: let \: ( - a, - b) = (h,k) \\ moreover \: its \: radius \: (r) = \sqrt{a {}^{2} - b {}^{2} }$

$so \: we \: know \: that \\ equation \: of \: circle \: of \: centre - radius \: form \: is \: given \: by \\ (x - h) {}^{2} + (y - k) {}^{2} = r {}^{2} \\ (x - ( - a)) {}^{2} + (y - ( - b)) {}^{2} = ( \sqrt{a {}^{2} - b {}^{2} } ) {}^{2} \\ (x + a) {}^{2} + (y + b) {}^{2} = a {}^{2} - b {}^{2} \\ x {}^{2} + a {}^{2} + 2ax + y {}^{2} + b {}^{2} + 2yb = a {}^{2} - b {}^{2} \\ ie \: \: x {}^{2} + y {}^{2} + 2ax + 2yb + b {}^{2} + b {}^{2} = 0 \\ x {}^{2} + y {}^{2} + 2b {}^{2} + 2ax + 2yb = 0$

$hence \: the \: equation \: of \: the \: circle \: with \: centre \: (-a, -b) \: and \: radius \: \sqrt{a {}^{2} - b {}^{2} } \: is \\ x {}^{2} + y {}^{2} + 2b {}^{2} + 2ax + 2yb = 0$