JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Circle

If the circles

x^{2} + y^{2} - 16 x - 20 y + 164 = r^{2}

and

(x - 4)^{2} + (y - 7)^{2} = 36

intersect at two distinct points, then:

(A) $0 < r < 1$
(B) $1 < r < 11$ ✓ Correct
(C) $r > 11$
(D) $r = 11$

Answer: (B)

Solution

Put both circles in centre--radius form. The first,

x^{2} + y^{2} - 16 x - 20 y + 164 = r^{2}

, becomes

(x - 8)^{2} + (y - 10)^{2} = r^{2}

, centre

(8, 10)

, radius

r

. The second,

(x - 4)^{2} + (y - 7)^{2} = 36

, has centre

(4, 7)

, radius

6

. The distance between centres is

4^{2} + 3^{2} = 5

. Two circles meet at two distinct points when

∣ r_{1} - r_{2} ∣ < d < r_{1} + r_{2}

, i.e.

∣ r - 6∣ < 5 < r + 6

. The right inequality holds for any positive

r

, and

∣ r - 6∣ < 5

gives

1 < r < 11

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option

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