How do you find the equation of the circle given Center (8,6), r=10?

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Answer 1 · 2016-07-14T11:38:56

$x^2+y^2-16x-12y=0$

In a circle, a point moves so that it always remains at a fixed distance (called radius) from a point (called center).

Let the point be $(x,y)$ and as center of circle is $(8,6)$ , its distance from this point will always be $10$ .

Hence equation of circle is given by

$(x-8)^2+(y-6)^2=10^2$ or

$x^2-16x+64+y^2-12y+36=100$ or

$x^2+y^2-16x-12y+64+36-100=0$ or

$x^2+y^2-16x-12y=0$