The circles x^2 + y^2 = 4 and \((x - 5)^2 + (y - 5)^2 = 40\) intersect in two points $A$ and $B.$ Find the slope of AB.
+128474 x^2 + y^2 = 4 (1)
(x - 5)^2 + (y - 5)^2 = 40 ⇒ x^2 + y^2 - 10x - 10y + 25 + 25 = 40 ⇒
4 - 10x - 10y + 50 = 40 ⇒ -10x -10y = -14 ⇒ 5x + 5y = 7 ⇒ 5y = 7 -5x ⇒ y = (7 -5x) / 5 (2)
Sub (2) into (1) for y
x^2 + [ (7- 5x)/5)^2 = 4
x^2 + [ 49 - 70x + 25x^2 ] / 25 = 4 multiply through by 25
25x^2 + 25x^2 - 70x + 49 = 100
50x^2 - 70x - 51 = 0
The approx solutions to this are x ≈ -.528 and x ≈ 1.92
And when x ≈ -.528 y ≈1.92
And when x ≈ 1.92 y ≈ -.528
The slope is [ 1.92 - - .528 ] / [ -.528 - 1.92] = [1.92 + .528 ] / - [ 1.92 + .528 ] = -1
