Question

The graphs of x^2+y^2+6x-24y+72=0 and x^2-y^2+6x+16y-46=0 intersect at 4 points.Compute the sum of distances of these four points

Answer 1

Answer:

Step-by-step explanation:

We don't have to find the coordinates of the four points.

Since

(x2+y2+6x−24y+72)−(x2−y2+6x+16y−46)=0

⇒2y2−40y+118=0

we can set the four points as (−3±α,β),(−3±γ,ω) where

β+ω=−(−40)/2=20,βω=118/2=59.

Now, noting that

(−3+α)2+β2+6(−3+α)−24β+72=0⇒α2+(β−2)2=20β−59

the sum of the distances can be represented as

2α2+(β−2)2−−−−−−−−−−−√+2γ2+(ω−2)2−−−−−−−−−−−√=2(20β−59−−−−−−−√+20ω−59−−−−−−−√)(1)

Then,

(20β−59−−−−−−−√+20ω−59−−−−−−−√)2=20(β+ω)−2⋅59+2202βω−20⋅59(β+ω)+592−−−−−−−−−−−−−−−−−−−−−−−√=20⋅20−2⋅59+2202⋅59−20⋅59⋅20+592−−−−−−−−−−−−−−−−−−−−−√=202

The result follows from (1).