If the circles \(x^{2}+y^{2}+6 x+8 y+16=0\) and \(x^{2}+y^{2}+2(3-\sqrt{3}) x+2(4-\sqrt{6}) y =\mathrm{k}+6 \sqrt{3}+8 \sqrt{6}, \mathrm{k}>0\), touch internally at the point \(P(\alpha, \beta)\), then \((\alpha+\sqrt{3})^{2}+(\beta+\sqrt{6})^{2}\) is equal to

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Circle

If the circles

x^{2} + y^{2} + 6 x + 8 y + 16 = 0

and

x^{2} + y^{2} + 2 (3 - 3) x + 2 (4 - 6) y = k + 63 + 86, k > 0

, touch internally at the point

P (α, β)

, then

(α + 3)^{2} + (β + 6)^{2}

is equal to

Answer: 25

The first circle has centre

C_{1} (- 3, - 4)

and radius

r_{1} = 9 + 16 - 16 = 3

. For the second, the centre is

C_{2} (3 - 3, 6 - 4)

and, once the

63

and

86

terms cancel, its radius satisfies

r_{2}^{2} = 34 + k

. The distance between the centres is

(3)^{2} + (6)^{2} = 9 = 3

. Internal contact requires this distance to equal

r_{2} - r_{1}

, so

34 + k - 3 = 3

, giving

34 + k = 6

and

k = 2 > 0

. The point of contact lies on the line of centres at distance

r_{1} = 3

from

C_{1}

, on the side away from

C_{2}

, namely

P = C_{1} + \frac{r _{1}}{∣ C _{1} C _{2} ∣} (C_{1} - C_{2}) = (- 3 - 3, - 4 - 6)

. Hence

(α + 3)^{2} + (β + 6)^{2} = (- 3)^{2} + (- 4)^{2} = 25

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Integer answer