# Using the following equation, find the center and radius: x^{2} + 4x + y^{2} - 6y = -4

**Solution:**

Given equation x^{2} + 4x + y^{2} - 6y = -4

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

In order to bring the given equation in standard form, let us convert the LHS into two perfect square trinomials.

We use completing the square method for this.

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

(x^{2} + 4x + 2^{2} ) + (y^{2} - 6y + 3^{2}) = -4 + 4 + 9

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

[since a^{2} + 2ab + b^{2 }= (a + b)^{2} and a^{2} - 2ab + b^{2 }= (a - b)^{2}]

(x + 2)^{2} + (y - 3)^{2} = 9

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

By comparing, we get centre (h, k) as (-2, 3) and radius r as 3.

## Using the following equation, find the center and radius: x^{2} + 4x + y^{2} - 6y = -4

**Summary:**

Using the following equation, x^{2} + 4x + y^{2} - 6y = -4 the centre is (-2, 3) and the radius is 3 units.

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