SOLUTION: Find the equation of the circle which passes through the origin and cuts off intercepts 3 and 4 from the positive parts of the axes respectively.
Question 1051841: Find the equation of the circle which passes through the origin and cuts off intercepts 3 and 4 from the positive parts of the axes respectively. Found 2 solutions by ikleyn, josgarithmetic:Answer by ikleyn(52792) (Show Source): You can put this solution on YOUR website! .
Find the equation of the circle which passes through the origin and cuts off intercepts 3 and 4 from the positive parts of the axes respectively.
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There are two such circles and, correspondingly, two equations.
The first circle has the center at (1.5,2) and the radius = = = 2.5:
= 6.25.
The second circle has the center at (2,1.5) and the same radius:
= 6.25.
Try finding any two lines, their perpendicular segment bisectors of their defining segments, and where those bisectors intersect. THAT would be the center of the circle. The final completion would use standard form for center (h,k) to fill ; although still need too, to find the radius.
(3,0) to (0,0):
Looking half-way between this horizontal segment, the line you want to bisect is . Up to now, no restriction on y (until later maybe).
(0,0) to (0,4):
Segment is vertical. Intuition should tell you, the perpendicular bisecting line to be .
What point is the intersection of ?
Intersection is ( 3/2, 2 ), which will be the center point for the circle you are trying to find.
Now you have an equation for this circle, .
Using the Distance Formula, do you know how to find the radius, r, to finish the equation?