wiz-icon
MyQuestionIcon
MyQuestionIcon
3
You visited us 3 times! Enjoying our articles? Unlock Full Access!
Question

The equation of the circle whose radius is 5 and which touches the circle x2+y22x4y20=0 externally at the point (5, 5), is

A
x2+y218x16y120=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
x2+y218x16y+120=0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
x2+y2+18x+16y120=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
x2+y2+18x16y+120=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B x2+y218x16y+120=0
Let the centre of the required circle be (x1,y1) and the centre of given circle is (1, 2). Since radii of both circles are same, therefore, point of contact (5, 5) is the mid point of the line joining the centres of both circles. Hence x1=9 and y1=8. Hence the required equation is (x9)2+(y8)2=25

x2+y218x16y+120=0.


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