SOLUTION: The line with the equation y=mx is tangent to the circle with centre (-8,0) and radius 4 at the point P(x,y). a)find coordinates of P in terms of m b)Show that m = +- sqrt (3

Question 864894: The line with the equation y=mx is tangent to the circle with centre (-8,0) and radius 4 at the point P(x,y).
a)find coordinates of P in terms of m
b)Show that m = +- sqrt (3)/ 3 and hence find the coordinates of P
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
(x+8)^2+y^2=16,
(x+8)^2+y^2=16, y=mx
m = 1/sqrt(3)
m =-1/sqrt(3)
Multiply by Sqrt(3)/Sqrt(3) to get whole number in denominator
and get +-sqrt(3)/3
+-1/sqrt(3)=+-sqrt(3)/3
y=+-sqrt(3)x/3
x=-6, y =2sqrt(3)
P=(-6,2sqrt(3))
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