Question 1040236: Determine the equation of the circle containing the points (-1,-1) , (3,-1) with the line x+2y-3=0 passing through the center.
I REALLY NEED HELP PLEASE.
Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website!
The x-coordinate of the center obviously must be halfway between
the x-coordinates of the two given points, (-1,-1) , (3,-1), since
they have the same y-coodinates and are connected by a horizontal
line. Half-way between their x-coordinates, -1 and 3, is 1.
When you substitute 1 for x in the equation of the line you get
1. So the center of the circle is obviously (h,k) = (1,1).
All that's left to do is
1. to get the radius r by finding the distance
between that center and either of the given points, using the distance
formula.
2. And then to substitute for h, k and r in the standard equation of a
circle, which is
(x-h)² + (y-k)² = r²
Edwin
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