wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The equation of the circle passing through (1, 1) and the points of intersection of x2+y2+13x3y=0 and 2x2+2y2+4x7y25=0 is

A
2x2+2y2+25x13y30=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
4x2+4y2+30x13y25=0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
x2+y2+15x13y25=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
4x2+4y2+25x13y30=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 4x2+4y2+30x13y25=0
Circle through points of intersection of given two circles is
(x2+y2+13x3y)+λ(2x2+2y2+4x7y25)=0(1+2λ)x2+(1+2λ)y2+(13+4λ)x+(37λ)y25λ=0
As it passes through (1, 1),
1+2λ+1+2λ+13+4λ37λ25λ=024λ+12=0λ=12 Required circle is2x2+2y2+15x13y2252=0or 4x2+4y2+30x13y25=0

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Family of Circles
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon