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# Find the equation of a circle that has its center at (-4, 3) and has a radius of 5.

**Solution:**

Given: Center(h, k) = (-4, 3) and radius(r) = 5.

We know that the general equation of circle with center (h, k) and radius with r is (x - h)^{2} + (y - k)^{2} = r^{2}

Substitute the values of h, k, r in the standard form, we get

⇒ (x - (-4))^{2} + (y - 3)^{2 }= 5^{2}

⇒ (x + 4)^{2} + (y - 3)^{2} = 25.

We know that algebraic identites (a + b)^{2} = a^{2} + 2ab + b^{2} and (a - b)^{2} = a^{2} - 2ab + b^{2}

⇒ x^{2} + 8x +16 + y^{2} - 6y + 9 = 25

⇒ x^{2} + y^{2} + 8x - 6y = 0

Therefore, the equation of a circle that has its center (-4, 3) and has a radius of 5 is (x + 4)^{2} + (y - 3)^{2} = 25.

## Find the equation of a circle that has its center at (-4, 3) and has a radius of 5.

**Summary:**

The equation of a circle that has its center at (-4, 3) and has a radius of 5 is (x + 4)^{2} + (y - 3)^{2} = 25.

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