Question 1202864: Find the value of k for which y=kx-2 is a tangent to the curve y^2=10x-x^2 Found 2 solutions by ikleyn, greenestamps:Answer by ikleyn(52781) (Show Source):
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Find the value of k for which y=kx-2 is a tangent to the curve y^2=10x-x^2
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We look for intersection points of the circle y^2 = 10x-x^2 with a straight line
y = kx-2 and find "k" from the condition that two intersection points merge into one.
For it, we substitute y = kx-2 into the equation of the circle, and we get
(kx-2)^2 = 10x - x^2
k^2*x^2 - 4kx + 4 = 10x - x^2
(k^2+1)x^2 - (4k+10) + 4 = 0.
Next, we calculate the discriminant of this quadratic equation
d = b^2 - 4ac = [-(4k+10)]^2 - 16(k^2+1) = 16k^2 + 80k + 100 - 16k^2 - 16 = 80k + 84
and equate it to zero
80k + 84 = 0.
It gives us
80k = - 84, k = = .
ANSWER. k = = -1.05.
