If the line $a x+y=c$, touches both the curves $x^{2}+y^{2}=1$ and $y^{2}-4 \sqrt{2} x$, then $|c|$ is equal to

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Circle

If the line

a x + y = c

, touches both the curves

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

and

y^{2} - 42 x

, then

∣ c ∣

is equal to

(A) $\frac{1}{2}$
(B) $2$ ✓ Correct
(C) $\frac{1}{2}$
(D)2

Answer: (B)

Solution

A tangent to the parabola

y^{2} = 42 x

can be written

y = m x + \frac{2}{m}

. For the same line to touch the unit circle

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

, its distance from the origin must be

1

\frac{∣ 2 / m ∣}{1 + m ^{2}} = 1

, which tidies to

\frac{2}{m ^{2}} = 1 + m^{2}

, i.e.

m^{4} + m^{2} - 2 = 0

and so

m^{2} = 1

. With

m = \pm 1

the intercept

\frac{2}{m} = \pm 2

, giving

∣ c ∣ = 2

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option

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