+128474 1. The circles x^2+y^2 = 4 and (x-3)^2+(y-5)^2=25 intersect in two points, say P and Q. Find the slope of line segment PQ.
x^2 + y^2 = 4
(x -3)^2 + (y - 5)^2 = 25
Expand the second equation
x^2 - 6x +9 + y^2 - 10y + 25 = 25
x^2 + y^2 - 6x - 10y + 9 = 0
4 - 6x - 10y = - 9
13 = 6x + 10y
y = [ 13 - 6x ] / 10 sub this into x^2 + y^2 = 4
x^2 + [ 13 - 6x]^2 / 10^2 = 4
100x^2 + 36x^2 - 156x + 169 = 400
136x^2 - 156x - 231 = 0
Using the quadratic formula....we get two values for x
39 25√15 39 25√15
x = ___ + ______ and x = ___ - ______
68 68 68 68
When x = the first value [ because of the messy math... I used WolframAlpha to calculate these y values ]
65 15 √15
y = ___ - _______
68 68
And with the second value of x,
65 15√15
y = ___ + _______
68 68
We can ignore the first fractions and the denominators of the second when calculating the slope
[ 15√15 + 15√15] √15 [ 15 + 15] 30 -3
Slope of PQ = ________________ = ____________ = ___ = ___
[ -25√15 - 25√15 ] √15 [ -25 - 25] -50 5
