# How to find the standard form of the equation of the specified circle given that it passes through the origin and has its center at (-5, 4)?

Jun 5, 2018

${x}^{2} + {y}^{2} + 10 x - 8 y = 0$

#### Explanation:

the eqn of a circle centre$\left(a , b\right) \text{ }$and radius $r$

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

we need to find the radius

we can use Pythagoras

since the circle passes through the origin we have

${r}^{2} = {\left(0 - - 5\right)}^{2} + {\left(0 - 4\right)}^{2}$

${r}^{2} = 25 + 16 = 41$

$\therefore \text{ eqn } {\left(x - - 5\right)}^{2} + {\left(y - 4\right)}^{2} = 41$

${\left(x + 5\right)}^{2} + {\left(y - 4\right)}^{2} = 41$

${x}^{2} + 10 x + 25 + {y}^{2} - 8 y + 16 = 41$

$\implies {x}^{2} + {y}^{2} + 10 x - 8 y = 0$