Find equation of circle whose radius is 3 and center is origin??

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Answer 1 · 2018-02-22T03:13:06

$x^2+y^2=9$

The general equation of a circle is:

$(x-h)^2+(y-k)^2=r^2$

We want a circle with an a center $(h,k)$ centered at the origin.
Therefore $(h,k)=(0,0)$

We also want a radius $3$ so $r^2=(3)^2=9$

Given these values, we can substitute into our equation to get:

$(x-0)^2+(y-0)^2=9$

Simplify and we get the equation:

$x^2+y^2=9$

This is the equation we want and its graph looks like

graph{x^2+y^2=9 [-10, 10, -5, 5]}