JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Circle

If a line,

y = m x + c

is a tangent to the circle,

(x - 3)^{2} + y^{2} = 1

and it is perpendicular to a line

L_{1}

, where

L_{1}

is the tangent to the circle,

x^{2} + y^{2} = 1

at the point

(\frac{1}{2}, \frac{1}{2})

; then:

(A) $c^{2} + 7 c + 6 = 0$
(B) $c^{2} - 6 c + 7 = 0$
(C) $c^{2} + 6 c + 7 = 0$ ✓ Correct
(D) $c^{2} - 7 c + 6 = 0$

Answer: (C)

Solution

The tangent to

x^{2} + y^{2} = 1

(\frac{1}{2}, \frac{1}{2})

has slope

- 1

, so a line perpendicular to it has slope

1

and the form

y = x + c

. For

y = x + c

to touch

(x - 3)^{2} + y^{2} = 1

, the distance from the centre

(3, 0)

to the line

x - y + c = 0

must equal the radius

1

\frac{∣3 - 0 + c ∣}{2} = 1

, so

∣ c + 3∣ = 2

and

(c + 3)^{2} = 2

, which expands to

c^{2} + 6 c + 7 = 0

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Single correct option

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