Question 611948
<pre>
The equation of a circle is

(x - h)² + (y - k)² = r²

We will need the center(h,k) and the radius r.  Let's draw the
graph:

{{{drawing(400,4800/13,-3,23,-6,18, red(line(-3,0,23,0),line(0,-6,0,18)),
circle(10,6,10),locate(-3,6.5,"(0,6)"),locate(10,0,16),locate(10,6.5,"(h,k)"),circle(10,6,.1),locate(22.5,0,X),locate(0.3,18,Y) )}}}

We need the center and the radius. Since the circle is tangent
to the y-axis at (0,6), we know that a radius drawn at that point
will be horizontal.  So the y-coordinate, k, of the center is the
same as the y-coordinate of the point k = 6, so the center is 
(h,6).  We don't know h, and must find it. However we know that
h = r.  So the center (h,6) is really (r,6) So we draw the radius 
(in green) from (0,6) to the center (r,6):

{{{drawing(400,4800/13,-3,23,-6,18, red(line(-3,0,23,0),line(0,-6,0,18)),
circle(10,6,10),locate(-3,6.5,"(0,6)"),locate(10,0,16),green(line(0,6,10,6)),locate(5.5,6,r), locate(10,6.5,"(r,6)"),locate(22.5,0,X),locate(0.3,18,Y) )}}}

From the center (r,6) we draw a perpendicular to the x-axis. It is
6 units long because the y-coordinate of the center is 6. Also it
bisects the 16-unit chord, dividing it into two 8-unit segments:

{{{drawing(400,4800/13,-3,23,-6,18, red(line(-3,0,23,0),line(0,-6,0,18)),
circle(10,6,10),locate(-3,6.5,"(0,6)"),locate(6,0,8),green(line(0,6,10,6),line(10,6,10,0)), locate(10,6.5,"(r,6)"),locate(14,0,8),locate(10.3,3.5,6),locate(22.5,0,X),locate(0.3,18,Y) )}}}

Next we draw a radius from the center (r,6) to the left end of the chord:

{{{drawing(400,4800/13,-3,23,-6,18, red(line(-3,0,23,0),line(0,-6,0,18)),
circle(10,6,10),locate(-3,6.5,"(0,6)"),locate(6,0,8),green(line(0,6,10,6),line(10,6,10,0),line(2,0,10,6)), locate(10,6.5,"(r,6)"),locate(14,0,8),locate(10.3,3.5,6),locate(4.5,6,r),locate(5.2,3.7,r),locate(22.5,0,X),locate(0.3,18,Y) )}}}

Now we have a right triangle and we can use the Pytagorean theorem
to find r:

r² = 8² + 6²
r² = 64 + 36
r² = 100
 r = 10

So the figure is now:

{{{drawing(400,4800/13,-3,23,-6,18, red(line(-3,0,23,0),line(0,-6,0,18)),
circle(10,6,10),locate(-3,6.5,"(0,6)"),locate(6,0,8),green(line(0,6,10,6),line(10,6,10,0),line(2,0,10,6)), locate(10,6.5,"(10,6)"),locate(14,0,8),locate(10.3,3.5,6),locate(4.5,6,10),locate(4.8,3.7,10),locate(22.5,0,X),locate(0.3,18,Y) )}}}

And the equation of the circle,

(x - h)² + (y - k)² = r²

becomes,

(x - 10)² + (y - 6)² = 10²

or

(x - 10)² + (y - 6)² = 100

There is another possible solution. That's because we could
have drawn the circle tangent on the left side of the y-axis,
and the center would have been (-10,6) and the equation would
have been

(x + 10)² + (y - 6)² = 100

And the graph would be the exact mirror image of the one above.


{{{drawing(400,4800/13,0,26,-6,18, red(line(26,0,0,0),line(23,-6,23,18)),
circle(13,6,10),locate(23,6.5,"(0,6)"),locate(-4,0,8),green(line(23,6,13,6),line(13,6,13,0),line(21,0,13,6)), locate(9,6.5,"(-10,6)"),locate(16,0,8),locate(12,3.5,6),locate(18.6,6,10),locate(18.2,3.7,10),locate(0,0,X),locate(23,18,Y) )}}}

Edwin</pre>