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

A circle S passes through the point (0, 1) and is orthogonal to the circles (x−1)2+y2=16 and x2+y2=1 then,

A
Radius of S is 8
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
Radius of S is 7
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
Centre of S is (-7, 1)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
Center of S is (-8, 1)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct options are
B Radius of S is 7
C Centre of S is (-7, 1)
(a) The general equation of a circle is
x2+y2+2gx+2fy+c=0
Where, centre and radius are given by (g,f) and g2+f2c, respectively.

(b) If the two circles x2+y2+2g1x+2f1y+c1=0 and x2+y2+2g2x+2f2y+c2=0 are orthogonal, then 2g1g2+2f1f2=c1+c2.

Let the circle be x2+y2+2gx+2fy+c=0
It passes through (0, 1).
1+2f+c=0 ....(i)
Orthogonal with
x2+y22x15=02g(1)=c15 using property (b) stated above
c=152g ....(ii)
Similarly,
Orthogonal with x2+y21=0
therefore c=1 .....(iii)
Solving (i), (ii) and (iii), we get
g=7 and f=1
Centre is (g,f)=(7,1)
Radius = g2+f2c
=49+11=7


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