A circle S passes through the point (0, 1) and is orthogonal to the circles x - 1 2 + y 2 = 16 and x ⁡…

A circle S passes through the point (0, 1) and is orthogonal to the circles x-12+y2=16 and x⁡ 2 + y⁡ 2 = 1 . Then
  1. Radius of S is 8
  2. Radius of S is 7
  3. Centre of S is (-7, 1)
  4. Centre of S is (-8, 1)

Solution

Given circles
x2+y2-2x-15=0
x2+y2-1=0
Radical axis x+7=0 ......(i)
Centre of circle lies on (i)
Let the centre be -7, k
Let equation be x2+y2+14x-2ky+c=0
Orthogonallity gives
- 14=c-15 ⇒c=1 .......(ii)
0, 1→1-2k+1=0 ⇒k=1
Hence radius = 72+k2-c= 49+1-1=7
Alternate Solution
Given circles x2+y2-2x-15=0
x2+y2-1=0
Let equation of circle x2+y2+2gx+2fy+c=0
Circle passes through (0, 1)
⇒ 1+2f+c=0
Applying condition of orthogonality
-2g=c-15, 0=c-1
⇒ c=1, g=7, f= -1
r= 49+1-1=7; centre -7, 1

Asked in: JEE Advanced 2014 (Paper 1)

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