wiz-icon
MyQuestionIcon
MyQuestionIcon
5
You visited us 5 times! Enjoying our articles? Unlock Full Access!
Question

The locus of midpoint of the chord of contact of x2+y2=2 from the points on 3x+4y=10 is a circle whose centre is

A
(45,35)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(35,45)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(410,310)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(310,410)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D (310,410)
Let the midpoint be (h,k).
Equation of chord of contact is
T=S1hx+ky2=h2+k22hx+ky(h2+k2)=0 (1)

Consider any point (x1,y1) on 3x+4y=10
Equation of chord of contact is
T=0xx1+yy12=0 (2)
Comparing equation (1) and (2), we get
x1h=y1k=2h2+k2x1=2hh2+k2, y1=2kh2+k2
Now putting (x1,y1) in 3x+4y=10, so
6h+8k=10h2+10k2x2+y26x108y10=0

Hence, the centre is (310,410)

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