CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The equation of the circle which cuts orthogonally each of the three circles given below:
x2+y22x+3y7=0, x2+y2+5x5y+9=0 and x2+y2+7x9y+29=0

A
x2+y216x8y12=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
x2+y216x18y4=0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
x2+y2+16x18y+4=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
x2+y2+16x+18y+12=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B x2+y216x18y4=0
Let the required circle be
x2+y2+2gx+2fy+c=0 (1)

Since, it is orthogonal to three given circles respectively, therefore
2g×(1)+2f×32=c7
or 2g+3f=c7(2)
2g×52+2f×(52)=c+9
or 5g5f=c+9(3)
and 2g×72+2f×(92)=c+29
or 7g9f=c+29(4)
Solving equations (2),(3) and (4) we get
g=8,f=9 and c=4
Substituting the values of g, f, c in (1) then required circle is
x2+y216x18y4=0

Alternate Solution:
S1:x2+y22x+3y7=0S2:x2+y2+5x5y+9=0S3:x2+y2+7x9y+29=0
radical axis for S1 and S2
S2S1=07x8y+16=0(1)
radical axis for S2 and S3
S3S2=0x2y+10=0(2)
From (1) and (2) radical centre of the given circle will be
(8,9)
Hence equation of the required circle will be
(x8)2+(y9)2=((S1))2x2+y216x18y4=0

flag
Suggest Corrections
thumbs-up
10
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon