The equation of a circle is(x-3)squared plus (y+2)squared.The point (8,-2)is on the circle.What is the equation of the line that is tangent to the circle at (8,-2)?
y=8
x=8
x=3
y=3
(x-3)^2 + (y+2)^2 = r^2
(5)^2 + 0^2 = r^2
r = 5
center at (3,-2) sketch it
(8,-2) is at the right edge of the circle at x = 8 and tangent is vertical
so
x = 8