(b,c)Given circles

x 2 + y 2 - 2x -15 = 0

x 2 + y 2 - 1 = 0

Radical axis x + 7 = 0 … (1)

Centre of circle lies on (1)

Let the centre be (-7, k)

Let equation be x 2 + y 2 + 14x - 2ky + c = 0

Orthogonallity gives

- 14 = c - 15 c = 1 … (2)

(0, 1) 1 - 2k + 1 = 0 k = 1

Hence radius =

Alternate solution

Given circles x 2 + y 2 – 2x – 15 = 0

x 2 + y 2 – 1 = 0

Let equation of circle x 2 + y 2 + 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

; centre ( – 7, 1)

──────────────────────────────────────────────────────────────────────────────────────────