How do you find the equation of the circle with center at (4,-6) and passes through point (7,-2)?

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Answer 1 · 2017-03-02T09:10:37

Equation of a circle is $(x-4)^2+(y+6)^2=25$ or $x^2+y^2-8x+12y+27=0$

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 $(4,-6)$ is

$(x-4)^2+(y-(-6))^2=r^2$ or $(x-4)^2+(y+6)^2=r^2$

As it passes through $(7,-2)$ , we will have

$(7-4)^2+(-2+6)^2=r^2$

or $3^2+4^2=r^2$ and $r^2=16+9=25$

Hence, equation of circle is $(x-4)^2+(y+6)^2=25$

or $x^2+y^2-8x+12y+27=0$ and its radius is $sqrt25=5$
graph{x^2+y^2-8x+12y+27=0 [-8, 16, -12, 0]}