Equation of a circle, given three intercepts

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i'll not familiar with this type of problem so i wasn't sure how to start it. i could really use some help.
a circle passes through the origin and has intercepts equal to 1 and 2 on the x- and y-axes respectively. find the equation of the cirlce.
thanks in advance for your help
 
3ptcir4ys.gif


This is what the question is looking for.
You can see the center.
What is the radius?
 
pka said:
You can see the center.
But is a picture acceptable "proof" of this?

Another method:

The general equation of a circle is:

. . . . .x<sup>2</sup> + y<sup>2</sup> + Ax + By + C = 0

You are given that (0, 0), (1, 0), and (0, 2) are solutions of the above equation. Then:

. . . . .(0, 0): 0<sup>2</sup> + 0<sup>2</sup> + A(0) + B(0) + C = 0

So C = 0, and the equation for this circle may be simplified as:

. . . . .x<sup>2</sup> + y<sup>2</sup> + Ax + By = 0

Now plug in the other two points, and solve for A and B. This will give you the equation (in general form) for the circle. To find the center and radius, complete the square.

Eliz.
 
stapel said:
But is a picture acceptable "proof" of this?
Eliz.
No of course not! But I assume that surely the student knows to find the intersection of perpendicular bisectors of the segments determined by the three non-collinear points.
 
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