Question 573501
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 {{{y=9}}}, and {{{y=-1}}}. It also has the vertical tangent {{{x=0}}} (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 {{{y=-1}}} must meet at the center of the circle, halfway between the lines. That tells you that the center has {{{y=(9+(-1))/2=4}}}, 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 {{{x=0}}}, meaning that the center could be at {{{x=5}}} or at {{{x=-5}}}. However, since the circle's center is in the second quadrant, it must be {{{x=-5}}}.
{{{drawing(300,300,-12,3,-3,12,
grid(1),
green(circle(-5,4,5)),
blue(line(-12,9,3,9)), blue(line(-12,-1,3,-1)),
red(circle(-5,4,0.2)), red(line(-5,4,0,4)),
red(line(-5,4,-5,9)),red(line(-5,4,-5,-1))
)}}} Circle in green. Horizontal tangents in blue. In red: x-axis, y-axis, center of the circle, and radii.
{{{(x+5)^2+(y-4)^2=25}}}