Question 1148112: what is the equation of a circle passing through (12,1) and (2,-3) with center on the line 2x-5y+10=0
Answer by ikleyn(52803)   (Show Source):
You can put this solution on YOUR website! .
The locus of the points equidistant from two given points is the perpendicular bisector to the segment connecting these points.
So, you need to construct the perpendicular bisector to the segment connecting the given points P = (12,1) and Q = (2,-3),
and then find its intersection with the given straight line.
The midpoint between the two given points is (7,-1).
The segment connecting (12,1) and (2,-3) has the slope  =  =  .
Hence, the perpendicular line (perpendicular bisector) has the slope  .
The line with the slope passing through the point (7,-1) has the equation
y - (-1) =  , or y =  .
The intersection of the straight lines
2x - 5y = -10 (1) (the given line) and
y = (2) (the perpendicular bisector)
is (solve the system by substitution) the point (x,y) = (5,4).
So, the center of the circle is the point (5,4).
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The radius of the circle is the distance from the point (5,4) to the point (12,1):
r =  =  =  .
Therefore, the equation of this circle is
 = 58. ANSWER
Solved.
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