Question

Identify the equation of the circle that has its center at (9, 12) and passes through the origin.

Answer 1

Answer:  $(x-9)^2 + (y-12)^2 = 225$

This is the same as writing (x-9)^2 + (x-12)^2 = 225

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Explanation:

Any circle equation fits the template of $(x-h)^2 + (y-k)^2 = r^2$

The center is (9,12) which tells us the values of h and k in that exact order.

h = 9

k = 12

To find the radius r, we need to find the distance from the center (9,12) to a point on the circle. The only point we know on the circle is the origin (0,0).

Apply the distance formula to find the distance from (9,12) to (0,0)

$d = \sqrt{ (x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{ (9-0)^2+(12-0)^2}\\\\d = \sqrt{ (9)^2+(12)^2}\\\\d = \sqrt{ 81+144}\\\\d = \sqrt{ 225}\\\\d = 15$

The distance from (9,12) to (0,0) is 15 units. Therefore, r = 15

An alternative to finding this r value is to apply the pythagorean theorem. The distance formula is effectively a modified version of the pythagorean theorem.

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Since h = 9, k = 12 and r = 15, we can then say:

$(x-h)^2 + (y-k)^2 = r^2\\\\(x-9)^2 + (y-12)^2 = 15^2\\\\(x-9)^2 + (y-12)^2 = 225$

which is the equation of this circle.