Hi Aime.
You can solve this problem several ways, because you are given more information that you need.
You are trying to find the equation of a circle. The standard form equation of a circle is
(x  h)^{2} + (y  k)^{2} = r^{2}
where (h, k) is the circle's centerpoint and r is the radius.
Method 1:
Any three points on a plane define a unique circle, so you could write three simultaneous circle equations could solve for h, k, and r:
(1  h)^{2} + (3  k)^{2} = r^{2}
(6  h)^{2} + (3  k)^{2} = r^{2}
(0  h)^{2} + (5  k)^{2} = r^{2}
Method 2:
Since point 1 is tangent to the circle, the radius lines on the line perpendicular to the given line equation at that point. The slope of the given line 3x+4y15=0 is m=A/B = 3/4, so the slope of the radius to (1,3) is 4/3. That makes its equation 4x  3y + C = 0 and thus C = 3(3)  4(1) = 5. So the radial line is 4x  3y + 5 = 0. Solving for y we get y = (4/3)x + 5/3. Since we know the center (h, k) lies on this line, we know k = (4/3)h + 5/3. So we can substitute this into the equation of the circle passing through points 1 and 2:
(1  h)^{2} + ( ((4/3)h + 5/3)  3)^{2} = r^{2}
(6  h)^{2} + ( ((4/3)h + 5/3)  3)^{2} = r^{2}
and solve for h, then use that to find k and r.
Method 3 (easiest):
Look at the points: (1, 3) (6, 3) and (0, 5). The first two points have the same y value; can we take advantage of that? Sure! Look again at method 1, but just the first two equations:
(1  h)^{2} + (3  k)^{2} = r^{2}
(6  h)^{2} + (3  k)^{2} = r^{2}
If we use the "elimination method" for simultaneous system of equations, we eliminate  in just one step  two variables:
(1  h)^{2}  (6  h)^{2} = 0.
And we can solve for h right away.
Hope this helps,
Stephen La Rocque.
