Question

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

Answer 1

`(x+6)^2+(y+8)^2 = 100`
or
`(x+6)^2+(y-8)^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 `+-8`

Applying the general form fro a circle with center `(a,b)` and radius `r`:
`color(white)("XXX") (x-a)^2+(y-b)^2=r^2`
gives two possibilities:

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

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