How do you write an equation of a circle with center at (-3,6) and a radius with endpoint (0,6)?

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Answer 1 · 2016-12-22T22:32:26

Please see the explanation.

The standard Cartesian form for the equation of a circle is:

$(x - h)^2 + (y - k)^2 = r^2" [1]"$

Where $(x, y)$ is any point on the circle, $(h, k)$ is the center point, and r is the radius.

We are given that the center point is $(-3, 6)$ , therefore, we substitute -3 for h and 6 for k into equation [1] and label it equation [2]:

$(x - -3)^2 + (y - 6)^2 = r^2" [2]"$

To find the value of the radius substitute the given point, $(0, 6)$ , into the equation [2] and then solve for r:

$(0 - -3)^2 + (6 - 6)^2 = r^2$

$r^2 = 9$

$r = 3$

Substitute 3 for r into equation [2] and then label it equation [3]:

$(x - -3)^2 + (y - 6)^2 = 3^2" [3]"$

Equation [3] is the equation of the circle.