# x^{2} + y^{2} + 8x + 22y + 37 = 0 in standard form?

**Solution: **

Given: Equation is x^{2} + y^{2} + 8x + 22y + 37 = 0 --- (1)

Eq (1) is the equation of a circle

We know the standard form of the equation of a circle is (x - α)^{2} +(y -β)^{2} = r^{2} --- (2)

Where (α, β) is the center of the circle, 'r' is the radius of the circle and (x, y) is any point on the circle.

So we have to convert the equation 1 in the form of equation 2

⇒ x^{2} + y^{2} + 8x + 22y + 37 = 0

Combining like terms

⇒ (x^{2} + 8x ) + (y^{2} + 22y) + 37 = 0

Now for making the perfect square we have to add and subtract similar terms using completing the square.

⇒ (x^{2} + 2.4.x + 4^{2}) - 4^{2} + (y^{2} + 11^{2} + 2.11.y) - 11^{2} + 37 = 0

⇒ (x + 4)^{2} + (y + 11)^{2} = 11^{2} + 4^{2} - 37

⇒ (x + 4)^{2} + (y + 11)^{2} = 121 + 16 - 37

⇒ (x + 4)^{2 }+ (y + 11)^{2} = 100

⇒ (x + 4)^{2} + (y + 11)^{2} = 10^{2}

⇒ (x - (-4))^{2} + (y - (-11))^{2} = 10^{2} --- (3)

Now this eq(3) is in the form of eq(2) i.e., the general form of the equation of the circle

By comparing eq(2) and (3) we get

The center of the circle = (-4 , -11)

The radius of the circle = 10units.

## x^{2} + y^{2} + 8x + 22y + 37 = 0 in standard form?

**Summary: **

On converting the equation x^{2} + y^{2} + 8x + 22y + 37 = 0 in the standard form, we get (x - (-4))^{2} + (y - (-11))^{2} = 10^{2} and the center of the circle as (-4 , -11) and radius as 10 units.