How do you find the radius of a circle with equation $(x-5)^2+(y-3)^2=r^2$ that contains the point at (5,1)?

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Answer 1 · 2017-08-23T23:09:57

The radius will measure $2$ units.

Since $(x, y) = (5, 1)$ , we have:

$(5 - 5)^2 + (1 - 3)^2 = r^2$

$0^2 + (-2)^2 = r^2$

$4 = r^2$

$r = 2$

Here is the corresponding graph.

graph{(x - 5)^2 + (y -3)^2 = 2^2 [-10, 10, -5, 5]}

Hopefully this helps!

How do you find the radius of a circle with equation (x-5)^2+(y-3)^2=r^2(x-5)^2+(y-3)^2=r^2 that contains the point at (5,1)?