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?

1 Answer
May 17, 2018

#(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]}