# What is the radius of a circle whose equation is x^{2} + y^{2} + 8x - 6y + 21 = 0?

**Solution:**

Given circle equation x^{2} + y^{2} + 8x - 6y + 21=0

We have standard equation of circle,

x^{2} + y^{2 }- 2ax - 2by + (a^{2} + b^{2} - m^{2}) = 0 with centre(a, b) and radius “m”

To find radius, we extract the “m” term

m = √(a^{2} +b^{2} ) --- (a)

Compare the terms (a^{2} + b^{2} - m^{2}) = 21

-2ax = 8x ⇒ a = -4

-2by = -6 ⇒ b = 3

Put a,b values in eq(a)

Radius = m = √(a^{2} +b^{2} ) =√((-4)^{2} +3^{2})

= √(16 +9)

= √25 = 5

Radius = 5 units

## What is the radius of a circle whose equation is x^{2} + y^{2} + 8x - 6y + 21 = 0?

**Summary:**

The radius of a circle whose equation is x^{2} + y^{2} + 8x - 6y + 21 = 0 is 5units.