# What is the length of the radius and the coordinates of the center of the circle defined by the equation (x+7)^2+(y-3)^2=121?

Mar 14, 2017

The radius is $11 \left(14 - 3\right)$ and the coordinates of the centre is ($7 , 3$)

#### Explanation:

Opening up the equation,
${\left(x + 7\right)}^{2} + {\left(y - 3\right)}^{2} = 121$
${x}^{2} + 14 x + 49 + {y}^{2} - 6 y + 9 = 121$
${y}^{2} - 6 y = 63 - {x}^{2} + 14 x$

Find the x-intercepts, and the midpoint to find x-line of symmetry,
When $y = 0$,
${x}^{2} - 14 x - 63 = 0$
$x = 17.58300524 \mathmr{and} x = - 3.58300524$

$\frac{17.58300524 - 3.58300524}{2} = 7$

Find the highest and lowest point and midpoint,
When $x = 7$,
${y}^{2} - 6 y - 112 = 0$
$y = 14 \mathmr{and} y = - 8$

$\frac{14 - 8}{2} = 3$

Hence, the radius is $11 \left(14 - 3\right)$ and the coordinates of the centre is ($7 , 3$)