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#1
26-12-2007, 20:16
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 Find equation of circle with this information? Circle lies in 3rd quadrant and touches x,y axes.If tangent from pt.-4,2 to circle is 5, find equation circle? Circle in 3rd Quadrant. Tangent from point A (-4,2) to circle is 5.
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#2
26-12-2007, 20:17
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| Clearly the equation is of the form (x+a)^2 + (y+a)^2=a^2 Additionally we know that (-4+a)^2+(2+a)^2 = 5^2+a^2 I get this from the fact that there must be a right triangle joining center of the circle and (-4,2) or 16+a^2-8a+4+a^2+4a=25+a^2 -5+a^2-4a=0 a=(4+/-sqrt(16+20))/2 a is positive so = 5 (sqrt(36)+4)/2
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