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Which is the center and radius of a circle with the equation x^(2)+y^(2)-12x+10y+12=0?

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To find the center and the radius of the circle with its equation, we will use the standard form of a circle's equation, which is (x-h)² + (y-k)² = r², where the center is (h,k) and the radius r.

Now, let's compute the center first.

The formulas for the center (h,k) are
h = -C/(2A) where A is the coefficient of x² which is 1 and C is the coefficient of x which is -12. Substituting these values we get:
h = --12/(2*1) = 6.

Similarly, k = -D/(2B) where B is the coefficient of y² which is 1 and D is the coefficient of y which is 10. Substituting these values we get:
k = -10/(2*1) = -5.

So, the center of the circle is (6,-5).

Next, we need to find the radius. The formula for the radius is √(h²+k²-E) where E is the constant term in the equation which is 12. Substituting the values of h,k and E we get:
radius = √((6)²+(-5)²-12) = √(36+25-12) = √49 = 7.

So the radius of the circle is 7.

In conclusion, the center of the circle given by the equation x² + y² - 12x + 10y + 12 = 0 is at the point (6,-5) and has a radius of 7 units.

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