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What is the equation of a circle with the center (3,1) and radius of 4?

1 Answer

Jun 25, 2018

${(x - 3)}^{2} + {(y - 1)}^{2} = 16$ $(x-3)^2+(y-1)^2=16$

Explanation:

The format for the equation of a circle is

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

where $(h, k)$ $(h,k)$ is the center and $r$ $r$ is the radius. We can just plug in values and solve from here:

${(x - 3)}^{2} + {(y - 1)}^{2} = 4^{2}$ $(x-3)^2+(y-1)^2=4^2$
${(x - 3)}^{2} + {(y - 1)}^{2} = 16$ $(x-3)^2+(y-1)^2=16$

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