## Posts Tagged ‘equation in standard form circle’

## Problem 1121

Write the equation in standard form.

x^{2} + y^{2} + 4y – 45 = 0

**Solution:-**

Complete the square and then write the equation in standard form.

(x – h)^{2} + (y – k)^{2} = r^{2}

Regroup the terms.

(x^{2}) + (y^{2} + 4y) – 45 = 0

Since x is square, there is nothing needed.

Now complete the square for y^{2} + 4y.

To complete the square, add and subtract 4 inside the second set of parentheses.

x^{2} + (y^{2} + 4y + 4 – 4 ) – 45 = 0

Now remove the subtract term from within the parentheses.

x^{2} + (y^{2} + 4y + 4 – 4 ) – 45 = 0

Use the addition principle to get the constant terms on the right side of the equation.

x^{2} + (y^{2} + 4y + 4 ) = 49

Now writ the equation in standard form.

(x -h)^{2} + (y – k)^{2} = r^{2}

(x – 0)^{2} + (y + 2)^{2} = 49