How do you find the radius of a circle with equation ${(x - 2)}^{2} + {(y - 2)}^{2} = r^{2}$ $(x-2)^2+(y-2)^2=r^2$ that contains the point at (2,5)?

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Answer 1 · 2018-04-24T18:18:30

$3$ $3$ .

If the circle ${(x - 2)}^{2} + {(y - 2)}^{2} = r^{2}$ $(x-2)^2+(y-2)^2=r^2$ passes through the

point $(2, 5)$ $(2,5)$ , the co-ords. of $(2, 5)$ $(2,5)$ must satisfy the given eqn.

$:. (2-2)^2+(5-2)^2=r^2=9$ , giving, $r=3$ .

But, in the given eqn. of circle, $r$ denotes the radius.

Therefore, the radius of the circle in question is $3$ .

How do you find the radius of a circle with equation (x-2)^2+(y-2)^2=r^2(x−2)2+(y−2)2=r2(x-2)^2+(y-2)^2=r^2 that contains the point at (2,5)?