(b,d)

Let C 1 and C 2 be the centre of circle S 1 and S 2 respectively

Let ∠ C 2 QM = ∠ C 2 MQ = θ ⇒ ∠ QC 2 M = π – 2 θ

Now vc ∠ QC 2 M + ∠ PC 1 M = π ⇒ π – 2 θ + π – 2 φ = π ⇒ θ + φ = π /2

Now vc ∠ QMP = π – ∠ QMC 2 – ∠ PMC 1 = π – ( θ + φ ) = π – π /2 = π /2

hence locus equation of variable point M is (x + 2)(x –2) + (y – 7)(y + 5) = 0

but locus of M does not contains point P and Q because P is included when radius of S 1 is zero and circle S 2 becomes straight line which is impossible. Q is included when radius of S 2 is zero and circle S 1 becomes straight line which is also impossible.

so set E 1 does not contain point P(–2, 7) and Q(2, – 5)

Locus of mid-points of chords passing through (1, 1) is h + K – (1 + k) = h 2 + k 2 – 2K

⇒ h 2 + K 2 – 2K – h + 1 = 0 ⇒ x 2 + y 2 – x – 2y + 1 = 0

Now equation of line passing through P(–2, 7) and R(1, 1) is ⇒ y + 2x – 3 = 0

Let centre of x 2 + y 2 – 2y – 39 = 0 is C 3 (0, 1) ⇒ centre of locus of M is C 3 (0, 1)

Now foot of C 3 (0, 1) on line y + 2x – 3 = 0 is . which is mid-point of chord PR of circle
x 2 + y 2 – 2y – 39 = 0

But if P is not the part of locus of M then PQ is not the chord of locus of M.

So point does not lies in set E 2