SOLUTION: The line mx-y=0 is tangent to the circle X2+Y2-6X+2y+2=0 find the two values of m.

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Question 602433: The line mx-y=0 is tangent to the circle X2+Y2-6X+2y+2=0 find the two values of m.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
If the line mx-y=0 <--> y=mx is tangent to the circle, they have only one point in common. So, substituting
y=mx into the equation for the circle,
x%5E2%2By%5E2-6x%2B7y%2B2=0, there should be only one point that is a solytion.
That would mean only one pair (x,y), meaning only one value for x.
So, here we go
x%5E2%2B%28mx%29%5E2-6x%2B2%28mx%29%2B2=0 --> x%5E2%2Bm%5E2x%5E2-6x%2B2mx%2B2=0 --> %28m%5E2%2B1%29x%5E2%2B%282m-6%29x%2B2=0
If a quadratic equation ax%5E2%2Bbx%2Bc=0 has only one solution its discriminant must be zero:
b%5E2-4%2Aa%2Ac=0
For the equation with m,
a=m%5E2%2B1, b=2m-6 and c=2
The discriminant is
%282m-6%29%5E2-4%2A%28m%5E2%2B1%29%2A2=0 --> 4m%5E2-24m%2B36-8%28m%5E2%2B1%29=0 --> 4m%5E2-24m%2B36-8m%5E2-8=0 --> -4m%5E2-24m%2B28=0
Dividing both sides of the equal sign by (-4), we get as much friendlier equation:
m%5E2%2B6m-7=0
and factoring, we get
%28m%2B7%29%28m-1%29=0 with solutions highlight%28m=1%29 and highlight%28m=-7%29 .