Patrick D. answered • 04/23/17
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Patrick the Math Doctor
The same question is for what values of k will the line intersect the circle at only one point.
The line equation is y = 1/2X + 1/2K
Plugging this into the equation of the circle:
(x-1)^2 + (1/2X + 1/2K -2 )^2 = 4
(x-1)^2 + (1/2X + B )^2 = 4 where B = 1/2K-2
x^2 - 2X + 1 + 1/4X^2 + BX + B^2 = 4
5/4X^2 + (B-2)X + B^2-3 = 0
plugging this into the quadratic formula with a=5/4 , b = B-2, and c= B^2-3
x = [ (2-B) +/- square-root ( (B-2)^2 - 4(5/4)(B^2-3) ) ] / (5/2)
x = [ (2-B) +/- square-root ( (B^2 - 4B +4 - 5B^2 + 15 ] / (5/2)
x = [ (2-B) +/- square-root ( -4B^2 - 4B + 19) ] / (5/2)
In order to guarantee there is only one root, the expression under the square root sign must be zero.
-4B^2 - 4B + 19 = 0
4B^2 + 4B - 19 = 0
Running this through the quadratic formula with A=4, B=4 and C=-19 and simplifying
B = -1/2 +/- rad5 , that is negative one-half plus or minus the square-root of 5
Changing back to k, where B = 1/2K-2
1/2K = 2 = -1/2 +/- rad5
In the positive case: 1/2K - 2 = -1/2 + rad5
K = 3 + 2*rad5 <-- is obtained by multiplying everything by 2 and solving for K
Now, it's time to check to see that it works. The equation of the line is
y = 1/2X + 1/2( 3 + 2*rad5) = 1/2X + 3/2 + rad5
subtracting 2 from both sides:
y - 2 = 1/2X + (rad5 - 1/2)
(y-2)^2 = 1/4X^2 + (rad5-1/2)X + (5 - rad5 + 1/4)
Substituting this into the equation of the circle:
(x-1)^2 + (y-2)^2 = 4
(x-1)^2 + 1/4X^2 + (rad5-1/2)X + (5 - rad5 + 1/4) = 4
x^2 - 2X + 1 + 1/4X^2 + (rad5-1/2)X + (5 - rad5 + 1/4) = 4
5/4 X^2 + (rad5 - 5/2)X + 9/4 - rad5 = 0
Verifying the discriminant of this quadratic equation is zero will suffice.
A = 5/4, B = rad5-5/2, C = 9/4-rad5
B^2 - 4Ac = (rad5-5/2)^2 - 4(5/4)(9/4-rad5)
= 5 - 5rad5 + 25/4 - 5/(9/4 - rad5)
= 5 - 5rad5 + 25/4 - 45/4 + 5rad5
= 20/4 - 5rad5 + 25/4 - 45/4 + 5rad5
= 0
The check for the negative case is left for you.
Answers are:
K = 3 + 2*rad5
k = 5 - 2*rad5
