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Question Number 27579 by Mcfaithal last updated on 10/Jan/18

A circle with center (−3,1) passes through the point (3,1). Find it′s equation

$$\mathrm{A}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{center}\:\left(−\mathrm{3},\mathrm{1}\right) \\ $$$$\mathrm{passes}\:\mathrm{through}\:\mathrm{the}\:\mathrm{point}\: \\ $$$$\left(\mathrm{3},\mathrm{1}\right).\:\mathrm{Find}\:\mathrm{it}'\mathrm{s}\:\mathrm{equation} \\ $$

Answered by Joel578 last updated on 10/Jan/18

r = distance between (−3,1) and (3,1) = (√((3 + 3)^2 + (1 − 1)^2 )) = 6 (x − a)^2 + (y − b)^2 = r^2 (x + 3)^2 + (y − 1)^2 = 36 x^2 + y^2 + 6x − 2y − 26 = 0

$${r}\:=\:\mathrm{distance}\:\mathrm{between}\:\left(−\mathrm{3},\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{3},\mathrm{1}\right) \\ $$$$\:\:\:\:=\:\sqrt{\left(\mathrm{3}\:+\:\mathrm{3}\right)^{\mathrm{2}} \:+\:\left(\mathrm{1}\:−\:\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:=\:\mathrm{6} \\ $$$$ \\ $$$$\left({x}\:−\:{a}\right)^{\mathrm{2}} \:+\:\left({y}\:−\:{b}\right)^{\mathrm{2}} \:=\:{r}^{\mathrm{2}} \\ $$$$\left({x}\:+\:\mathrm{3}\right)^{\mathrm{2}} \:+\:\left({y}\:−\:\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{36} \\ $$$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:\mathrm{6}{x}\:−\:\mathrm{2}{y}\:−\:\mathrm{26}\:=\:\mathrm{0} \\ $$

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