JEE Main · 2011 · 10 Apr 2011 · Offline · Circle

The two circles

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

and

x^{2} + y^{2} = c^{2} (c > 0)

touch each other if

(A) $∣ a ∣ = c$ ✓ Correct
(B) $a = 2 c$
(C) $∣ a ∣ = 2 c$
(D) $2∣ a ∣ = c$

Answer: (A)

Solution

The first circle

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

(x - \frac{a}{2})^{2} + y^{2} = \frac{a ^{2}}{4}

, with centre

(\frac{a}{2}, 0)

and radius

\frac{∣ a ∣}{2}

; the second is centred at the origin with radius

c

. Both pass through the origin, and they touch when the distance between centres equals the difference of the radii,

\frac{∣ a ∣}{2} = c - \frac{∣ a ∣}{2}

, which gives

∣ a ∣ = c

JEE Main · 2011 · 10 Apr 2011 · Offline · Single correct option

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