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

The equation of the circle whose centre is on the line 2xy2=0 and passes through the points (3,2) and (2,0) is

A
x2+y2x+2y=6
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
x2+y2x2y=13
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
x2+y26x+2y=9
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
x2+y23x+2y=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A x2+y2x+2y=6
Let equation of circle be
(xh)2+(yk)2=r2
Putting the points (3,2) and (2,0), we get
(3h)2+(2k)2=r2 (1)
(2h)2+(0k)2=r2 (2)
From equation (1) and (2), we get
(3h)2+(2k)2=(2h)2+(0k)296h+4+4k=4+4h10h4k9=0 (3)

Also, the centre lies on line 2xy2=0, so
2hk=2 (4)
From equation (3) and (4), we get
(h,k)=(12,1)

From equation (2),
(212)2+(1)2=r2r2=294r=292

Hence, the equation of circle is
(x12)2+(y+1)2=294
x2+y2x+2y=6

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Definition of Circle
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon