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

1 Answer
Feb 24, 2016

Answer:

#(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]}