SOLUTION: Given the circle's equation which is (x+4)^2+(y+8)^2=136 and the point of tangency which is (2,2), find the slope of the tangent.
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-> SOLUTION: Given the circle's equation which is (x+4)^2+(y+8)^2=136 and the point of tangency which is (2,2), find the slope of the tangent.
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Question 1024055: Given the circle's equation which is (x+4)^2+(y+8)^2=136 and the point of tangency which is (2,2), find the slope of the tangent. Answer by Cromlix(4381) (Show Source):
You can put this solution on YOUR website! Hi there,
(x+4)^2+(y+8)^2=136
Centre of circle = (-4, -8)
Gradient = y2 - y1/x2 - x1
Gradient of line between:-
(-4, -8) and (2,2)
Gradient = 2 - (-8)/2 - (-4)
Gradient = 2 + 8/2 + 4
Gradient = 10/6 = 5/3
As tangent at right angles to
the line from the centre of the circle
The two gradients multiply together
to give -1
m1 x m2 = -1
5/3 x m2 = -1
m2 = -3/5
This is the slope of the tangent.
Hope this helps :-)
