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A-circle-with-center-3-1-passes-through-the-point-3-1-Find-it-s-equation-




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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