<PRE><B>Question 141251</B>
If the point(3,-2) lies on the circle defined by x^2+y^2+mx-4y-3=0, what is the value of m???
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Substitute for x and for y to solve for m:
3^2 + (-2)^2 +3m -4*-2 -3 = 0

9 + 4 + 3m + 8 -3 = 0

3m = -18
 m = -6
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Cheers,
Stan H.</PRE>