   SEARCH HOME Math Central Quandaries & Queries  Question from aime, a student: Find the equation of the circle tangent to 3x+4y-15=0 at P1(1,3) and passing through P2(6,3) and P3(0,5)? 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 = r2
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 = r2
(6 - h)2 + (3 - k)2 = r2
(0 - h)2 + (5 - k)2 = r2

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+4y-15=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 = r2
(6 - h)2 + ( ((4/3)h + 5/3) - 3)2 = r2

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 = r2
(6 - h)2 + (3 - k)2 = r2

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.     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.