# Write the equation of the circle with center (2, -3) and a radius of 4.

**Solution:**

The equation of a circle is

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

Where (h, k) is the center

r is the radius

It is given that

Center = (2, -3)

Radius = 4

Substituting the values in the formula

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

(x - 2)^{2} + (y + 3)^{2} = 16

Expanding using the algebraic identity

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

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

We get

x^{2} + 4 - 4x + y^{2} + 9 + 6y = 16

x^{2} + y^{2} - 4x + 6y = 16 - 13

x^{2} + y^{2} - 4x + 6y = 3

Therefore, the equation of the circle is x^{2} + y^{2} - 4x + 6y = 3.

## Write the equation of the circle with center (2, -3) and a radius of 4.

**Summary:**

The equation of the circle with center (2, -3) and a radius of 4 is x^{2} + y^{2} - 4x + 6y = 3.

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