SOLUTION: Find the values of k for which the line y=kx-10 meets the curve x^2+y^2=10x.

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Question 1184256: Find the values of k for which the line y=kx-10 meets the curve x^2+y^2=10x.
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!



x%5E2%2By%5E2=10x....eq.1
y=kx-10+.....eq.2
--------------------------------
substitute y in eq.1
x%5E2%2B%28kx-10%29%5E2=10x
x%5E2%2Bk%5E2x%5E2+-+20kx+-10x%2B+100=0
%281%2Bk%5E2%29x%5E2+-+%2820k+%2B10%29x%2B+100=0

if the line meets the curve, there is one solution and it is a case if discriminant =0

so, use discriminant
b%5E2-4ac=0 ...in your case a=%281%2Bk%5E2%29, b=-+%2820k+%2B10%29, and c=100
%28-+%2820k+%2B10%29%29%5E2-4%281%2Bk%5E2%29%2A100=0++
400k%5E2+%2B+400k+%2B+100-400+k%5E2+-+400=0+

400k+-+300=0++
+400k+=+300+

4k+=+3
k+=+3%2F4
and your line is:
y=%283%2F4%29x-10+






Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.


            In her post,  @MathLover1  INCORRECTLY  interprets the problem,
            considering only the case when the straight line touches the circle.

            The correct interpretation is to consider a general case of an intersection
            ("having at least one intersection point").

            So,  I  came to bring a  CORRECT  SOLUTION.


x%5E2%2By%5E2=10x....eq.1

y=kx-10+.....eq.2
--------------------------------

substitute y in eq.1

x%5E2%2B%28kx-10%29%5E2+=+10x

x%5E2%2Bk%5E2x%5E2+-+20kx+-10x%2B+100=0

%281%2Bk%5E2%29x%5E2+-+%2820k+%2B10%29x%2B+100=0


if the line meets the curve, there is at least one solution and it is a case if discriminant >= 0


so, use discriminant 

b%5E2-4ac+%3E=+0  ...in your case a=%281%2Bk%5E2%29, b=-+%2820k+%2B10%29, and c=100

%28-+%2820k+%2B10%29%29%5E2-4%281%2Bk%5E2%29%2A100+%3E=+0++

400k%5E2+%2B+400k+%2B+100-400+k%5E2+-+400+%3E=+0+
 
 400k+-+300+%3E=+0++

+400k+%3E=+300+
 
4k+%3E=+3

k+%3E=+3%2F4


ANSWER.  k >= 3%2F4.

Solved (correctly).