Sunday

April 20, 2014

April 20, 2014

Posted by **Melba** on Tuesday, September 26, 2006 at 3:48pm.

Any ideas??

First express the line as in point slope form

y=(5/2)x - 21/2

The center must be on the line, so it is a point expressed as

(x,(5/2)x - 21/2)

In order for the circle to be tangent to both axis the center must be the same distance from both axis, so

x=y or x=(5/2)x - 21/2 solve for x

The standard equation for the circle is

(x-a)^2+(y-b)^2=r^2

You should be able to determine a,b and r.

After review, I don't know if I suggested how to find all the solutions.

You need to look at |x|=|y| or

|x|=|5/2)x - 21/2|

There should be one more solution to the question.

Thanks for all your help...

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