Question 1124597
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Draw a rough sketch showing the line and the two given points.<br>
An arbitrary point on the line has coordinates (x,2x-3).  We need the distance from (x,2x-3) to (5,2) to be the same as the distance from (x,2x-3) to (3,-2).<br>
If the distances are equal, then the squares of the distances are equal.  So, to avoid equations involving several square roots, we can write and solve an equation that says the squares of the distances are equal.  That equation would be<br>
{{{(x-5)^2+(2x-5)^2 = (x-3)^2+(2x-1)^2}}}<br>
But before we go down that computational path, take a look at your sketch.  If you have drawn it well, it should look as if the point (2,1) is the point on the line that is equidistant from the two given points.<br>
And checking the distances from (2,1) to each of (5,2) and (3,-2), we see that indeed the distances are the same.<br>
So we don't need to do the ugly algebra; the center of the circle is (2,1).<br>
The distance from (2,1) to each of the two given points is sqrt(10); so we have all we need to write the equation of the circle.<br>
Answer:  The equation of the circle is
{{{(x-2)^2+(y-1)^2 = 10}}}<br>