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

If two circles x2+y22ax+c2=0 and x2+y22by+c2=0 touch each other externally, then

A
1a2+1c2=1b2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1a2+1b2=1c2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
1a2+1b2=2c2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1b2+1c2=1a2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 1a2+1b2=1c2
Let S1:x2+y22ax+c2=0
C1(a,0) and r1=a2c2
and S2:x2+y22by+c2=0
C2(0,b) and r2=b2c2

C1C2=(a0)2+(0b)2
C1C2=a2+b2
and r1+r2=a2c2+b2c2
Since the two circles touch each other externally, we have
C1C2=r1+r2
a2+b2=a2c2+b2c2
a2+b2=a2c2+b2c2+2a2c2b2c2
c2=a2c2b2c2
c4=a2b2c2(a2+b2)+c4
a2b2=c2(a2+b2)
1a2+1b2=1c2

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon