What is the standard form of the equation of a circle with a center(1,-2) and passes through (6,-6)?

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Answer 1 · 2018-07-09T08:02:36

The circle equation in standard form is

${(x - x_{0})}_{0}^{2} + {(y - y_{0})}_{0}^{2} = r^{2}$ $(x-x_0)^2+(y-y_0)^2=r^2$

Where $(x_{0}, y_{0}); r$ $(x_0,y_0); r$ are the center coordinates and radius

We know that $(x_{0}, y_{0}) = (1, - 2)$ $(x_0,y_0)=(1,-2)$ , then

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

But we know that passes trough $(6, - 6)$ $(6,-6)$ , then

${(6 - 1)}^{2} + {(- 6 + 2)}^{2} = r^{2}$ $(6-1)^2+(-6+2)^2=r^2$

$5^{2} + {(- 4)}^{2} = 41 = r^{2}$ $5^2+(-4)^2=41=r^2$ , So $r = \sqrt{41}$ $r=sqrt41$

Finally we have the standard form of this circle

${(x - 1)}^{2} + {(y + 2)}^{2} = 41$ $(x-1)^2+(y+2)^2=41$ .

Answer 2 · 2018-07-09T08:07:12

${(x - 1)}^{2} + {(y + 2)}^{2} = 41$ $(x-1)^2+(y+2)^2=41$

Let the equation of unknown circle with center $(x_{1}, y_{1}) \equiv (1, - 2)$ $(x_1, y_1)\equiv(1, -2)$ & radius $r$ $r$ be as follows

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

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

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

Since, the above circle passes through the point $(6, - 6)$ $(6, -6)$ hence it will satisfy the equation of circle as follows

${(6 - 1)}^{2} + {(- 6 + 2)}^{2} = r^{2}$ $(6-1)^2+(-6+2)^2=r^2$

$r^{2} = 25 + 16 = 41$ $r^2=25+16=41$

setting $r^{2} = 41$ $r^2=41$ , we get the equation of circle

${(x - 1)}^{2} + {(y + 2)}^{2} = 41$ $(x-1)^2+(y+2)^2=41$