# What are the center and radius of the circle defined by the equation x^2+y^2-6x+8y+21=0?

Dec 23, 2016

The center is $\left(3 , - 4\right)$ and the radius is $= 2$

#### Explanation:

We need ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

Let's rearrange the equation as

${x}^{2} - 6 x + {y}^{2} + 8 y = - 21$

We can now complete the squares

${x}^{2} - 6 x + 9 + {y}^{2} + 8 y + 16 = - 21 + 9 + 16$

We can now factorise

${\left(x - 3\right)}^{2} + {\left(y + 4\right)}^{2} = 4 = {2}^{2}$

This is the equaton of a circle, center $\left(3 , - 4\right)$ and radius $= 2$

graph{(x-3)^2+(y+4)^2-4=0 [-6.42, 9.38, -7.07, 0.83]}

Dec 23, 2016

Center is $\left(3 , - 4\right)$ and radius is $2$.

#### Explanation:

${x}^{2} + {y}^{2} - 6 x + 8 y + 21 = 0$ can be written as

${x}^{2} - 6 x + {y}^{2} + 8 y = - 21$

or ${x}^{2} - 2 \times 3 \times x + {3}^{2} + {y}^{2} + 2 \times 4 \times y + {4}^{2} = - 21 + {3}^{2} + {4}^{2}$

or ${\left(x - 3\right)}^{2} + {\left(y + 4\right)}^{2} = - 21 + 9 + 16$

or ${\left(x - 3\right)}^{2} + {\left(y + 4\right)}^{2} = {2}^{2}$

This is the locus of a point which moves so that its distance from point $\left(3 , - 4\right)$ is always $2$.

Hence this is the equation of a circle whose center is $\left(3 , - 4\right)$ and radius is $2$.
graph{x^2+y^2-6x+8y+21=0 [-2.043, 7.957, -6.12, -1.12]}