The equation of a circle is (x-3)^2+(y+2)^2=25. The point (8,-2) is on the circle.

What is the equation of the line that is the tangent to the circle at (8,-2)?

y=8
x=8
x=3
y=3

Question

The equation of a circle is (x-3)^2+(y+2)^2=25. The point (8,-2) is on the circle.

What is the equation of the line that is the tangent to the circle at (8,-2)?

y=8
x=8
x=3
y=3

jim766 · Answer

do you know what the center of the circle is?

jim766 · Answer

you can tell by looking at the equation, do you know how?

anonymous · Answer

Nope

jim766 · Answer

(x-3)^2 +(y +2)^2 = 25         to find the center point, take the opposite of 
                                               what is in each parentheses

   (3, -2)     is the center,  do you see where I got that?

anonymous · Answer

Yea I do

jim766 · Answer

|dw:1370636234574:dw|

that means the point the gave us will be on the right side

jim766 · Answer

the answer will be a vertical line that goes through 8|dw:1370636446731:dw|