SOLUTION: The centre of a circle is on the line y=2x-3. The circle also passes through A(5,2) and B(3,-2). Determine the equation of the circle.

Question 1124597: The centre of a circle is on the line y=2x-3. The circle also passes through A(5,2) and B(3,-2). Determine the equation of the circle.
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20855) (Show Source): You can put this solution on YOUR website!
a circle is on the line and also passes through A(,) and B(,)
graph a line and given points:

|
|
|->->->

find coordinates of the midpoint of A(,) and B(,)
(,)
(,)...use it to find equation of a line perpendicular to the line -> slope is
perpendicular lines have slopes negative reciprocal to each other
so, our perpendicular line, , will have a slope
...to find plug in coordinates of the midpoint (,)

line is:

let's add it to the graph

the point where lines and should intersect in one point which is the center o the circle

-------------------solve system; since left sides are same, we have
...solve or

...both sides multiply by

now find

so, intersection point is: C(,)=> and coordinates o the center
add to graph

now we need radius
since circle also passes through A(,) and B(,), use one of these points and find distance between that point and center

C(,)
A(,)

add circle to the graph

and, your equation of the circle is:
where and coordinates o the center, is radius
plug in values:C(,)=> and and

Answer by greenestamps(13334) (Show Source): You can put this solution on YOUR website!

Draw a rough sketch showing the line and the two given points.

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).

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

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.

And checking the distances from (2,1) to each of (5,2) and (3,-2), we see that indeed the distances are the same.

So we don't need to do the ugly algebra; the center of the circle is (2,1).

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.

Answer: The equation of the circle is

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