Find the equation of the circle having (7,9) and (11,-7) as end of it'

1.	Find the equation of the circle having (7,9) and (11,-7) as end of it's diameter
Answer» ANSWER: HEY here is your SOLUTION PLS mark it as brainliest Step-by-step EXPLANATION: $so \: here \: for \: a \: certain \: circle \\ let \: (7,9) = (x1,y1) \\ (11, - 7) = (x2,y2)$ $so \: we \: know \: that \\ equation \: of \: diameter \: of \: circle \: is \: given \: by \\ (x - x1)(x - x2) + (y - y1)(y - y2) = 0 \\ ie \: (x - 7)(x - 11) + (y - 9)(y - ( - 7)) = 0 \\ (x - 7)(x - 11) + (y - 9)(y + 7) = 0 \\ \\ x {}^{2} - 11x - 7x + 77 + y {}^{2} + 7y - 9y - 63 = 0 \\ x {}^{2} - 18x + y {}^{2} - 2y + 14 = 0 \\ ie \: \: x {}^{2} + y {}^{2} - 18x - 2y + 14 = 0 \\ \\ thus \: our \: required \: equation \: of \: diameter \: is \: \\ x {}^{2} + y {}^{2} - 18x - 2y + 14 = 0$

Find the equation of the circle having (7,9) and (11,-7) as end of it's diameter

Answer»

PLS mark it as brainliest

Step-by-step EXPLANATION:

$so \: here \: for \: a \: certain \: circle \\ let \: (7,9) = (x1,y1) \\ (11, - 7) = (x2,y2)$

$so \: we \: know \: that \\ equation \: of \: diameter \: of \: circle \: is \: given \: by \\ (x - x1)(x - x2) + (y - y1)(y - y2) = 0 \\ ie \: (x - 7)(x - 11) + (y - 9)(y - ( - 7)) = 0 \\ (x - 7)(x - 11) + (y - 9)(y + 7) = 0 \\ \\ x {}^{2} - 11x - 7x + 77 + y {}^{2} + 7y - 9y - 63 = 0 \\ x {}^{2} - 18x + y {}^{2} - 2y + 14 = 0 \\ ie \: \: x {}^{2} + y {}^{2} - 18x - 2y + 14 = 0 \\ \\ thus \: our \: required \: equation \: of \: diameter \: is \: \\ x {}^{2} + y {}^{2} - 18x - 2y + 14 = 0$

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