SOLUTION: find equation of the circle passing through A(1,4) , B(-1,8) and tangent to the line x + 3y - 3 = 0

There are obviously two solutions.



We set the three radii equal. We use the distance formula
for two of them and the formula for the perpendicular
distance from a point to a line for the radius to the
point of tangency.





Square all three 



Taking the first two



That simplifies down to 



Substitute 2k-12 for h in the second and third





That simplifies to 



So k = 5 or k = 13

Substituting in



h = -2 or h = 14

the centers of the two circles are (-2,5) and (14,13) 

To find the radius, substitute in any one of the 
three expressions for the radius, say, the third one







The standard equation of a circle is:



So the smaller circle has equation



For the larger circle:







So the larger circle has equation



Edwin

The center of the circle lies on the perpendicular bisector to the segment with endpoints A and B.

The midpoint is C = (,) = (0,6).

The segment AB has the slope  =  = -2; hence, the perpendicular line has the slope  = 0.5.

The equation of the straight line with the slope 0.5 passing through (0,6) is

y-6 = 0.5*(x-0),   or  y = 0.5x + 6.   (1)


Now we need to find the center as the point (x,y) which lies at the straight line (1) and is equidistant from the point A and 
from the given straight line.

The distance from the point (x,y) to A is .

The distance the point (x,y) to the line x + 3y -3 = 0 is

 =      (see the lesson The distance from a point to a straight line in a coordinate plane in this site).

So, your equation is 

 = ,

or, when squared,

 = .       (2)

The second equation is

y = 0.5x + 6.                  (3)

When you substitute (3) into (2), you will get a single equation for x:

 = .


Simplify it step by step:

 =   ====>   =   ====>   =   ====>  

 =   ====>  =   ====>   = 0  ====>   = 0  ====>

 = .


There are two solutions:  = 14  and   = -2.


The corresponding centers are  (x,y) = (14,13)  and  (x,y) = (-2,5).


The distance from  (14,13)  to  (1,4)  is   =  = .

The distance from  (-2,5)  to  (1,4)  is   = .


So, there are two circles, and their equations are

 = 250   and    = 10.