What is the standard form of the equation of a circle with r = 5; (h, k) = (-5, 2)?

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What is the standard form of the equation of a circle with r = 5; (h, k) = (-5, 2)?

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Answer 1 · 2015-12-09T22:45:13

${(x + 5)}^{2} + {(y - 2)}^{2} = 25$ $(x+5)^2+(y-2)^2=25$

Explanation:

The standard form of the equation of a circle of radius $r$ $r$ centered at the point $(h, k)$ $(h,k)$ is ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ $(x-h)^2+(y-k)^2=r^2$ .

This equation is reflecting the fact that such a circle consists of all points in the plane that are distance $r$ $r$ from $(h, k)$ $(h,k)$ . If a point $P$ $P$ has rectangular coordinates $(x, y)$ $(x,y)$ , then the distance between $P$ $P$ and $(h, k)$ $(h,k)$ is given by the distance formula $\sqrt{{(x - h)}^{2} + {(y - k)}^{2}}$ $sqrt{(x-h)^2+(y-k)^2}$ (which itself comes from the Pythagorean Theorem).

Setting that equal to $r$ $r$ and squaring both sides gives the equation ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ $(x-h)^2+(y-k)^2=r^2$ .

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What is the standard form of the equation of a circle with r = 5; (h, k) = (-5, 2)?

1 Answer

Explanation:

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