How do you find the equation of the circle with center at (-3, 1) and through the point (2, 13)?

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Answer 1 · 2018-07-09T08:42:26

${(x - 1)}^{2} + {(y + 2)}^{2} = 169$ $(x-1)^2+(y+2)^2=169$

Let the equation of unknown circle with center $(x_{1}, y_{1}) \equiv (- 3, 1)$ $(x_1, y_1)\equiv(-3, 1)$ & radius $r$ $r$ be as follows

${(x - x_{1})}_{1}^{2} + {(y - y_{1})}_{1}^{2} = r^{2}$ $(x-x_1)^2+(y-y_1)^2=r^2$

${(x - (- 3))}^{2} + {(y - 1)}^{2} = r^{2}$ $(x-(-3))^2+(y-1)^2=r^2$

${(x + 3)}^{2} + {(y - 1)}^{2} = r^{2}$ $(x+3)^2+(y-1)^2=r^2$

Since, the above circle passes through the point $(2, 13)$ $(2, 13)$ hence it will satisfy the equation of circle as follows

${(2 + 3)}^{2} + {(13 - 1)}^{2} = r^{2}$ $(2+3)^2+(13-1)^2=r^2$

$r^{2} = 25 + 144 = 169$ $r^2=25+144=169$

setting $r^{2} = 169$ $r^2=169$ , we get the equation of circle

${(x - 1)}^{2} + {(y + 2)}^{2} = 169$ $(x-1)^2+(y+2)^2=169$