# How do you find the center and radius of the circle:  x^2 + y^2 – 10x + 6y + 18 = 0?

Mar 23, 2018

Centre is $\left(5 , - 3\right)$ and the Radius is $4$

#### Explanation:

We must write this equation in the form ${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$
Where $\left(a , b\right)$ are the co ordinates of the center of the circle and the radius is $r$.

So the equation is ${x}^{2} + {y}^{2} - 10 x + 6 y + 18 = 0$

Complete the squares so add 25 on both sides of the equation

${x}^{2} + {y}^{2} - 10 x + 25 + 6 y + 18 = 0 + 25$

= ${\left(x - 5\right)}^{2} + {y}^{2} + 6 y + 18 = 0 + 25$

Now add 9 on both sides

${\left(x - 5\right)}^{2} + {y}^{2} + 6 y + 18 + 9 = 0 + 25 + 9$
=${\left(x - 5\right)}^{2} + {\left(y + 3\right)}^{2} + 18 = 0 + 25 + 9$

This becomes

${\left(x - 5\right)}^{2} + {\left(y + 3\right)}^{2} = 16$

So we can see that the centre is $\left(5 , - 3\right)$ and the radius is $\sqrt{16}$ or 4

Mar 23, 2018

centre : $C \left(5 , - 3\right)$
radius : $r = 4$

#### Explanation:

The general equation of a circle:

color(red)(x^2+y^2+2gx+2fy+c=0...........to(1),

whose centre is color(red)(C((-g,-f)) and radius is color(red)(r=sqrt(g^2+f^2-c)

We have,

${x}^{2} + {y}^{2} - 10 x + 6 y + 18 = 0$

Comparing with $e q {u}^{n} \left(1\right)$, we get

$2 g = - 10 , 2 f = 6 \mathmr{and} c = 18$

$\implies g = - 5 , f = 3 \mathmr{and} c = 18$

So,

radius $r = \sqrt{{\left(- 5\right)}^{2} + {\left(3\right)}^{2} - 18} = \sqrt{25 + 9 - 18} = \sqrt{16} = 4$

i.e. $r = 4 > 0$

centre $C \left(- g , - f\right) \implies C \left(- \left(- 5\right) , - 3\right)$

i.e. centre $C \left(5 , - 3\right)$