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# Enter the equation of the circle described below. center (3, -2), radius = 5

**Solution:**

Given,

Center (3, -2), radius = 5

We know that the equation of circle with centre (h,k) and radius r units is

(x - h)^{2} + (y - k)^{2} = r^{2}

The equation of the circle with centre (3, -2), radius = 5 is

(x - 3)^{2} + (y + 2)^{2} = 5^{2}

Using the algebraic identities

(a - b)^{2} = a^{2} + b^{2} - 2ab

(a + b)^{2 }= a^{2 }+ b^{2} + 2ab

x^{2} - 6x + 9 + y^{2} +4y + 4 = 25

x^{2} + y^{2 }- 6x + 4y + 13 = 25

So we get

x^{2} + y^{2} - 6x + 4y + 13 - 25 = 0

x^{2} + y^{2} - 6x + 4y - 12 = 0.

Therefore, the equation of the circle is x^{2} + y^{2} - 6x + 4y - 12 = 0.

## Enter the equation of the circle described below. center (3, -2), radius = 5

**Summary: **

The equation of the circle with centre (3, -2), radius = 5 is x^{2} + y^{2} - 6x + 4y - 12 = 0.

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