CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Equation of the circle, centred at (1,5) and touching the line 3x+4y=8, is

A
x2+y22x+10y+1=0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
x2+y24x+5y+25=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
x2+y2+2x10y+26=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
x2+y210x+2y+1=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A x2+y22x+10y+1=0
Let r be the radius of required circle.
Then r is perpendicular distance of point (1,5) from 3x+4y8=0
r=|3(1)+4(5)8|32+42r=5
Hence required equation of circle is,
(x1)2+(y+5)2=52x2+y22x+10y+1=0

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Circle and Point on the Plane
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon