SOLUTION: The Problem says to find an equation of a circle that is tangent to both axes, has it's center in the second quadrant, & has a radius of 3. Thank you for your help!!!

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Question 24811: The Problem says to find an equation of a circle that is tangent to both axes, has it's center in the second quadrant, & has a radius of 3. Thank you for your help!!!
Answer by Earlsdon(6098) About Me  (Show Source):
You can put this solution on YOUR website!
Starting with the standard form of the equation for a circle with centre at (h, k) and radius, r:
%28x+-+h%29%5E2+%2B+%28y+-+h%29%5E2+=+r%5E2
You are given the radius, so r+=+3 and r%5E2+=+9
The circle is tangent to both the x-axis and the y-axis in quadrant II, so by inspection you can see that the centre is located at (-3, 3) so h = -3 and k = 3.
The equation therefore is:
%28x+%2B+3%29%5E2+%2B+%28y+-+3%29%5E2+=+9