SOLUTION: My math teacher gave me a problem with the following instructions: "Write the equation for the circle with center in the second quadrant; tangent to y=-1, y=9, and the y-axis." A

Question 573501: My math teacher gave me a problem with the following instructions:
"Write the equation for the circle with center in the second quadrant; tangent to y=-1, y=9, and the y-axis."
At first I tried to just plot y=-1 and y=9 to have an idea of where the circle might be, but I can't seem to solve the problem with the information that I have been provided. If it is not too much to ask, could I please get the solution to this specific problem and a detailed, step-by-step explanation of what to do when I encounter something like this? Any answer is much appreciated.
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
When a circle is tangent to a line, it touches that line at just one point. If you draw the radius from that point to the center of the circle, that radius is perpendicular to the line. The circle in the problem is tangent to horizontal lines , and . It also has the vertical tangent (the y-axis).
For the tangent circle in the problem, the radius from the point of intersection with the radius from the point of intersection with must meet at the center of the circle, halfway between the lines. That tells you that the center has , and is at distance 5 from either horizontal line.
The center of that circle will also be at distance 5 from the y-axis, which is , meaning that the center could be at or at . However, since the circle's center is in the second quadrant, it must be .
Circle in green. Horizontal tangents in blue. In red: x-axis, y-axis, center of the circle, and radii.

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