What is the equation of the circle with a center at $(6, 3)$ $(6 ,3 )$ and a radius of $1$ $1$ ?

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What is the equation of the circle with a center at $(6, 3)$ $(6 ,3 )$ and a radius of $1$ $1$ ?

1 Answer

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Answer 1 · 2016-02-03T18:51:20

${(x - 6)}^{2} + {(y - 3)}^{2} = 1$ $(x-6)^2+(y-3)^2=1$

Explanation:

The general equation of a circle with radius $r$ $r$ and center at $(h, k)$ $(h,k)$ is ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ $(x-h)^2+(y-k)^2=r^2$ . Since the center of this circle is at $(6, 3)$ $(6,3)$ and the radius is $1$ $1$ , we substitute to get:
${(x - (6))}^{2} + {(y - (3))}^{2} = {(1)}^{2}$ $(x-(6))^2+(y-(3))^2=(1)^2$

Finish it off for the final result:
${(x - 6)}^{2} + {(y - 3)}^{2} = 1$ $(x-6)^2+(y-3)^2=1$

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What is the equation of the circle with a center at $(6, 3)$ $(6 ,3 )$ and a radius of $1$ $1$ ?

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