Question

If the straight line x-y=k and the circle x^2+y^2+2x-4y-1=0 intersect at A and B, then the x-coordinate of the mid-point of AB is _ ?
A. 1+k
B. 1-k
C. (1+k)/2
D. (1-k)/2

Answer 1

since y=x-k,
x^2 + (x-k)^2 + 2x - 4(x-k) - 1 = 0
2x^2 - 2x(k+1) + k^2+4k-1 = 0
x = [2(k+1)±√(4(k+1)^2-8(k^2+4k-1))]/4

now, that looks like a lot of complicated stuff, but notice that it's really

(k+1)/2 ± some stuff

So, (k+1)/2 is the midpoint coordinate