JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Circle

Let the centre of a circle C be

(α, β)

and its radius

r < 8

. Let

3 x + 4 y = 24

and

3 x - 4 y = 32

be two tangents and

4 x + 3 y = 1

be a normal to C. Then

(α - β + r)

is equal to

(A)7✓ Correct
(B)9
(C)5
(D)6

Answer: (A)

Solution

Since

4 x + 3 y = 1

is a normal, the centre

(α, β)

lies on it, so

4 α + 3 β = 1

. The centre is equidistant from the two tangents, giving

\frac{3 α + 4 β - 24}{5} = \frac{3 α - 4 β - 32}{5}

. Taking

3 α + 4 β - 24 = 3 α - 4 β - 32

yields

8 β = - 8

, so

β = - 1

, and then

4 α - 3 = 1

gives

α = 1

. The radius is

r = \frac{∣3 ( 1 ) + 4 ( - 1 ) - 24∣}{5} = \frac{25}{5} = 5

, which satisfies

r < 8

. The other bisector leads to

α = \frac{28}{3}

with

r = \frac{80}{9} > 8

, which is rejected. Therefore

α - β + r = 1 - (- 1) + 5 = 7

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Single correct option

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