Question
Mon January 24, 2011 By: Puneet Singh

The equation of circle passing through (1,1) and point's of intersection of x^2+y^2+13x-3y=0 and 2x^2+2y^2+4x-7y-25=0.

Expert Reply
Tue February 22, 2011
Dear Student
 
The equation of circle passing through point  of intersection of two circles is
(x2+y2+13x-3y)+k (2x2+2y2+4x-7y-25)=0 ....(I)
Now circle passes through (1,1) so
1+1+13-3+k(2+2+4-7-25)=0
So 12+k(-24) =0 or k =-1/2
Substituting k =-1/2 in (I) required equation can be obtained.
Regards
Team Topper
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