Question

Find the equation to the circle described on the common chord of the circles `x^2+y^2-4x-2y-31=0` and `2x^2+2y^2-6x+8y-35=0` as diameter?

Answer 1

`(x-1.41)^2+(y+2.49)^2=23.385`

Explanation:

For this we should first find points of intersection of the two circles

`x^2+y^2-4x-2y-31=0` .............(A)

and `2x^2+2y^2-6x+8y-35=0` .............(B)

Doubling (A) and subtracting from (B), we get

`-6x+8x+8y+4y-35+62=0` or `2x+12y=-27`

or `x=-(27+12y)/2`

Putting this in (A) we get

`(27+12y)^2/4+y^2+4(27+12y)/2-2y-31=0`

`36y^2+162y+729/4+y^2+54+24y-2y-31=0`

or `37y^2+184y+821/4=0`

or `y=(-184+-sqrt(184^2-37xx821))/74`

= `(-184+-sqrt3479)/74=(-184+-58.98)/74`

i.e. `-3.28` or `-1.69`

and `x=-(27-12xx3.28)/2=6.18` and `-(27-12xx1.69)/2=-3.36`

Hence, points of intersection are `(6.18,-3.28)` and `(-3.36,-1.69)`

and center of circle would be `((6.18-3.36)/2,(-3.28-1.69)/2)` or `(1.41,-2.49)`

and radius would be half the distance between two points of intersection i.e. `1/2sqrt((6.18+3.36)^2+(-3.28+1.69)^2)`

= `1/2sqrt(91.0116+2.5281)=1/2sqrt93.5397=9.672/2=4.836`

and equation of circle is `(x-1.41)^2+(y+2.49)^2=23.385`

graph{(x^2+y^2-4x-2y-31)(2x^2+2y^2-6x+8y-35)(2x+12y+27)((x-1.41)^2+(y+2.49)^2-23.385)=0 [-9.75, 10.25, -6.68, 3.32]}