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

$(x-x_0)^2+(y-y_0)^2=r^2$

Where $(x_0,y_0); r$ are the center coordinates and radius

We know that $(x_0,y_0)=(1,-2)$ , then

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

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

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

$5^2+(-4)^2=41=r^2$ , So $r=sqrt41$

Finally we have the standard form of this circle

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

Answer 2 · 2018-07-09T08:07:12

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

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

$(x-x_1)^2+(y-y_1)^2=r^2$

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

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

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

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

$r^2=25+16=41$

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

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