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November 25, 2014

November 25, 2014

Posted by **ceasar1** on Wednesday, August 3, 2011 at 7:56am.

- mathematics -
**Reiny**, Wednesday, August 3, 2011 at 9:12amIf the circle is tangent to both axes and passes through (-3,6), I see its centre to be in the 2nd quadrant.

If the radius is r, then its centre must be (-r,r) and its equation is

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

x^2 + 2rx + r^2 + y^2 - 2ry + r^2 = r^2

at (-3,6)

9 - 6r + r^2 + 36 - 12r = 0

r^2 - 18r + 45 = 0

(r-3)(r-15) = 0

r = 3 or r = 15

since you want the smaller circle ... r = 3

equation:

(x+3)^2 + (y-3)^2 = 9

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