How do you find the equation of a circle with center at the point (-8,5) and tangent to the x-axis?

1 Answer
Jan 16, 2016

Answer:

#(x+8)^2+(y-5)^2=5^2#

Explanation:

A circle with center #(x_c,y_c)# tangent to the x-axis has a radius, #r#, equal to the distance from the center to the x-axis, namely #y_c#.

The general formula for a circle with center #(x_c,y_c)# and radius #r# is
#color(white)("XXX")(x-x_c)^2+(y-y_c)^2=r^2#

For the given values #(x_c,y_c)=(-8,5)#
and derived value #r=5#
this gives us:
#color(white)("XXX")(x+8)^2+(y-5)^2=5^2#
graph{(x+8)^2+(y-5)^2=25 [-16.86, 5.64, -1.03, 10.22]}