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

Two circles x2+y2+2ax+c=0 and x2+y2+2by=0 touch if

A
a2+b2=c2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1c=1a2+1b2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
1c2=1a2+1b2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1c2=1a2+1b2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is D 1c=1a2+1b2
s1=x2+y2+2ax+c=0
c1=(a,0)r1=a2c
S2=x2+y2+2by+c=0
c2=(0,b), r2=a2c
Since the two order touch each other,
r1+r2=c1c2
a2c+b2c=(a0)2+(b0)2
Squaring the above , we get
(a2c+b2c)2=(a)2+(b)2
a2c+b2c+2a2cb2c=a2+b2
2c+2a2cb2c=0
c=a2cb2c
c2=(a2c)(b2c)
a2b2a2ccb2+c2=c2
a2b2=a2c+b2c
a2b2a2+b2=c
1c=a2+b2a2b2=1a2+1b2
1c=1a2+1b2 answer
Option (b) is correct

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Touching Circles Theorem
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon