JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Circle

The circle passing through the point

(- 1, 0)

and touching the

y

-axis at

(0, 2)

also passes through the point

(A) $(- \frac{3}{2}, 0)$
(B) $(- \frac{5}{2}, 2)$
(C) $(- \frac{3}{2}, \frac{5}{2})$
(D)(-4, 0)✓ Correct

Answer: (D)

Solution

A circle touching the

y

-axis at

(0, 2)

has the form

(x - h)^{2} + (y - 2)^{2} = h^{2}

, where the center is

(h, 2)

and radius is

∣ h ∣

.

Alternatively, the equation is

x^{2} + (y - 2)^{2} + λ x = 0

.

Since it passes through

(- 1, 0)

1 + 4 - λ = 0 ⟹ λ = 5

So the circle is

x^{2} + y^{2} + 5 x - 4 y + 4 = 0

.

Setting

y = 0

x^{2} + 5 x + 4 = 0

, giving

(x + 1) (x + 4) = 0

, so

x = - 1

x = - 4

.

Therefore, the circle passes through

(- 4, 0)

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Single correct option

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