How do you write the equation of the circle with the given center at (-2,4) and passes through the point at (1,-7)?

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Answer 1 · 2017-03-02T09:52:54

Equation of the circle is $(x+2)^2+(y-4)^2=121$ or $x^2+y^2+4x-8y-101=0$

Equation of a circle with center at $(h,k)$ is

$(x-h)^2+(y-k)^2=r^2$ , where $r$ is its radius.

Hence equation of a circle with center at $(-2,4)$ is

$(x-(-2))^2+(y-4)^2=r^2$ or $(x+2)^2+(y-4)^2=r^2$

As it passes through $(1,-7)$ , we will have

$(-2+2)^2+(-7-4)^2=r^2$

or $0^2+11^2=r^2$ and $r^2=0+121=121$

Hence, equation of circle is $(x+2)^2+(y-4)^2=121$

or $x^2+y^2+4x-8y-101=0$ and its radius is $sqrt121=11$
graph{x^2+y^2+4x-8y-101=0 [-28, 24, -10, 16]}