Pre-cal question help!?

Question

A circle has center (-2,-2) and radius 3. Find all values of y such that the point (1,y) lies on the circle.

I HATE CIRCLES, assistance please! I'll mark as the best answer!

Answer 1

only (1,-2) would be on the circle, so y= {-2}

Answer 2

Picture the graph of your circle. The circle is centered in the 3rd quadrant (lower left) at -2,-2; the circle has radius of 3, so it just extends to the right to x=-2+3 = 1.

You're looking for points along the line x=1; the only one is that tangent point where the circle reaches that line, which would be at the single point 1, -2.

Answer 3

simplest thing to do is to draw the circle from the information you are given, though you should be able to solve it with basic equations. If you draw it then you will see the only point at which the circle center (-2,-2) radius 3 reaches a point where X = 1 is at the point (1,-2)

Answer 4

Circles generally have an equation of this form:

(x-a)^2 + (y-b)^2 = r^2 where the radius is r and (a,b) is the center. In your case, the equation is (x+2)^2 + (y+2)^2 = 9. Just enter in x=1 and solve for y.

(1+2)^2 + (y+2)^2 = 9
(3)^2 + (y+2)^2 = 9
9 + (y+2)^2 = 9
(y+2)^2 = 0
y=-2

Answer 5

First, draw a picture to help visualize the problem. Draw a point at (-2,-2) its center; (-5,-2) its leftmost point; (-2,-5) its lowest point; (-2,1) its highest point; and (1,-2) its rightmost point. Then draw a rough circle intersecting all 4 of the boundary points.
Notice anything?
The only place where the circle gets to the x=1 line is its rightmost point, (1,-2), where the circle is tangent (only touching at one point) the x=1 line.
So, the only place that meets the requirements is (1,-2), so y=-2, that's it.