Question 827899: Find he equation of all tangents to the circle x^2 + y^2 - 2x + 8y - 23 = 0
a) at the point (3,-10) on it
b) having slope of 3
Thanks so much in advance:)
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Find he equation of all tangents to the circle x^2 + y^2 - 2x + 8y - 23 = 0
a) at the point (3,-10) on it
b) having slope of 3
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Find the standard eqn of the circle.
x^2 + y^2 - 2x + 8y - 23 = 0
(x-1)^2 + (y+4)^2 = 40
Its center is (1,-4) and radius sqrt(40)
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a) at the point (3,-10) on it
Use a circle about the Origin with radius sqrt(40)
--> x^2 + y^2 = 40
The tangent and any point is -x/y
Point (3,-10) is translated to (2,-6)
Slope = 1/3
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Eqn of a line thru (3,-10) with a slope of 3 is
y+10 = (1/3)*(x-3)
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b) having slope of 3
Using the circle about the Origin:
-x/y = 3
x^2 + y^2 = 40
x = -3y
9y^2 + y^2 = 40
y = -2, +2
--> (6,-2)
Translated to the given circle:
--> (7,-6)
y+6 = 3(x-7)
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--> (-6,2)
Translate to the given circle
--> (-5,-2)
y+2 = 3x+15
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