Question
write the equation in standard form for the circle passing through (12,0) centered at the origin
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First, we can find the radius of the circle using the distance formula:
r = sqrt((12-0)^2 + (0-0)^2) = 12
Next, we can use the standard form of the equation for a circle:
(x - h)^2 + (y - k)^2 = r^2
where (h,k) is the center of the circle and r is the radius.
In this case, the center is at the origin (0,0), so we can simplify the equation to:
x^2 + y^2 = r^2
Substituting r = 12, we get:
x^2 + y^2 = 144
So the standard form of the equation for the circle passing through (12,0) centered at the origin is:
x^2 + y^2 = 144
r = sqrt((12-0)^2 + (0-0)^2) = 12
Next, we can use the standard form of the equation for a circle:
(x - h)^2 + (y - k)^2 = r^2
where (h,k) is the center of the circle and r is the radius.
In this case, the center is at the origin (0,0), so we can simplify the equation to:
x^2 + y^2 = r^2
Substituting r = 12, we get:
x^2 + y^2 = 144
So the standard form of the equation for the circle passing through (12,0) centered at the origin is:
x^2 + y^2 = 144
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