JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Circle

Let

r_{1}

and

r_{2}

be the radii of the largest and smallest circles, respectively, which pass through the point

(- 4, 1)

and having their centres on the circumference of the circle

x^{2} + y^{2} + 2 x + 4 y - 4 = 0

. If

\frac{r _{1}}{r _{2}} = a + b 2

, then

a + b

is equal to:

(A)5✓ Correct
(B)11
(C)7
(D)3

Answer: (A)

Solution

The circle

x^{2} + y^{2} + 2 x + 4 y - 4 = 0

has centre

(- 1, - 2)

and radius

1 + 4 + 4 = 3

. A circle of the family has centre

C

on this circumference and passes through

A (- 4, 1)

, so its radius is

C A

. The distance from

(- 1, - 2)

(- 4, 1)

(- 3)^{2} + 3^{2} = 32 > 3

A

lies outside the circle and

C A

varies between

32 - 3

and

32 + 3

. Hence

\frac{r _{1}}{r _{2}} = \frac{3 2 + 3}{3 2 - 3} = \frac{2 + 1}{2 - 1} = (2 + 1)^{2} = 3 + 22

a = 3

b = 2

and

a + b = 5

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Single correct option

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