# How do you write the equation of the circle with Center at (4, 4); passing through (7, 14)?

Oct 19, 2016

Equation of circle is ${x}^{2} + {y}^{2} - 8 x - 8 y - 77 = 0$

#### Explanation:

As the circle has center at $\left(4 , 4\right)$ and passes through $\left(7 , 14\right)$,

distance between a point $\left(x , y\right)$ and $\left(4 , 4\right)$ and between $\left(7 , 14\right)$ and $\left(4 , 4\right)$ will be equal. Hence,

$\sqrt{{\left(x - 4\right)}^{2} + {\left(y - 4\right)}^{2}} = \sqrt{{\left(7 - 4\right)}^{2} + {\left(14 - 4\right)}^{2}}$

or squaring ${\left(x - 4\right)}^{2} + {\left(y - 4\right)}^{2} = {\left(7 - 4\right)}^{2} + {\left(14 - 4\right)}^{2}$

or ${x}^{2} - 8 x + 16 + {y}^{2} - 8 y + 16 = {3}^{2} + {10}^{2}$

or ${x}^{2} + {y}^{2} - 8 x - 8 y + 32 - 9 - 100 = 0$

or ${x}^{2} + {y}^{2} - 8 x - 8 y - 77 = 0$