Question 1202864
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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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<pre>
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 = {{{-84/80}}} = {{{-21/20}}}.


<U>ANSWER</U>.  k = {{{-21/20}}} = -1.05.
</pre>

Solved.