# How do you write the standard equation for the circle passes through the origin, radius=10 and abcissa of the center is -6?

Feb 24, 2016

${\left(x + 6\right)}^{2} + {\left(y + 8\right)}^{2} = 100$
or
${\left(x + 6\right)}^{2} + {\left(y - 8\right)}^{2} = 100$

#### Explanation:

The "abcissa" in this question is equivalent to the $x$ coordinate.
There are two possible $y$ coordinate values that satisfy the given restrictions.

If the radius is $10$ then the distance from the origin to the center is $10$
and if the $x$ coordinate is $- 6$ then (by the Pythagorean Theorem) the $y$ coordinate is $\pm 8$

Applying the general form fro a circle with center $\left(a , b\right)$ and radius $r$:
$\textcolor{w h i t e}{\text{XXX}} {\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$
gives two possibilities:

The graph for ${\left(x + 6\right)}^{2} + {\left(y + 8\right)}^{2} = 100$
graph{(x+6)^2+(y+8)^2=100 [-27.37, 23.96, -19.48, 6.17]}

The graph for ${\left(x + 6\right)}^{2} + {\left(y - 8\right)}^{2} = 100$
graph{(x+6)^2+(y-8)^2=100 [-31.95, 19.37, -5.64, 20.02]}