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14 Aug 2006, 04:32
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If the circle with center (4,0) has a radius of 4, and the line through P and T has an equation of y=6-mx, then what is the value of m?
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14 Aug 2006, 05:28
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Eqn for circle: (x-4)^2 + y^2 = 16
Eqn for line: y = -mx + 6
Once I equate them together for the point of tengency, I get three variables (x, y m)? Then what?
Eqn for line: y = -mx + 6
Once I equate them together for the point of tengency, I get three variables (x, y m)? Then what?
14 Aug 2006, 10:27
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There are two relationships here
1) the point on the line lies on the circle so x, 6-xm lies on the circle
(x-4)^2 +(6-mx)^2 = 16
x^2 - 8x +16 + 36+ (mx)^2 -12mx = 16
X^2(1+m^2) -4x(2+3m) +36 =0 ---(1)
2) the slope of the tangent at T is perpendicular to the slope of the line joining the center and point T
(6-mx)/(x-4) = 1/m
6m - m^2x = x -4
x( 1+ m^2) -6m -4 = 0
x= 2(2+3m)/ (1+m^2)
now substitute this value of X in equation 1...
4(2+3m)^2/ (1+m^2) -8(2+3m)^2/ (1+m^2) +36 =0
(ideally at this stage you should be able to plug in any of the values of m given in the q)
(2+3m)^2 - 2(2+3m)^2 +9(1+m^2) =0
(2+3m)^2 = 9 (1+m^2)
4+ 9m^2 +12 m = 9 + 9 m^2
12m =5
so m = 5/12
OA??
1) the point on the line lies on the circle so x, 6-xm lies on the circle
(x-4)^2 +(6-mx)^2 = 16
x^2 - 8x +16 + 36+ (mx)^2 -12mx = 16
X^2(1+m^2) -4x(2+3m) +36 =0 ---(1)
2) the slope of the tangent at T is perpendicular to the slope of the line joining the center and point T
(6-mx)/(x-4) = 1/m
6m - m^2x = x -4
x( 1+ m^2) -6m -4 = 0
x= 2(2+3m)/ (1+m^2)
now substitute this value of X in equation 1...
4(2+3m)^2/ (1+m^2) -8(2+3m)^2/ (1+m^2) +36 =0
(ideally at this stage you should be able to plug in any of the values of m given in the q)
(2+3m)^2 - 2(2+3m)^2 +9(1+m^2) =0
(2+3m)^2 = 9 (1+m^2)
4+ 9m^2 +12 m = 9 + 9 m^2
12m =5
so m = 5/12
OA??
