SOLUTION: find the length of the tangent (-4,1) on the circle given by the equation x^2+y^2+8x+12y+5=0.

Question 906935: find the length of the tangent (-4,1) on the circle given by the equation x^2+y^2+8x+12y+5=0.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
find the length of the tangent (-4,1) on the circle given by the equation x^2+y^2+8x+12y+5=0.
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Find the center of the circle.
Put the equation is standard form.
(x+4)^2 + (y+6)^2 = 47
--> center at (-4,-6)
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Move the circle and the point 6 units up and 4 units right.
Now the circle is x^2 + y^2 = 47 and the point is (0,7)
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Find the tangent that passes thru (0,7)
(x,y) is a point on the circle
The slope of the tangent m1 = (y-7)/x
The slope of the tangent at any point on the circle m2 = -x/y
m1 = m2
(y-7)/x = -x/y
-x^2 = y^2 - 7y
x^2 + y^2 = 7y
x^2 + y^2 = 47
7y = 47
y = 47/7
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One tangent point is (sqrt(94)/7,(47/7))
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Find the distance between those 2 points.
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