JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Circle

If a circle

C

passing through the point

(4, 0)

touches the circle

x^{2} + y^{2} + 4 x - 6 y = 12

externally at the point

(1, - 1)

, then the radius of

C

is:

(A) $25$
(B)4
(C)5✓ Correct
(D) $57$

Answer: (C)

Solution

The point

(1, - 1)

lies on the given circle

x^{2} + y^{2} + 4 x - 6 y - 12 = 0

, and the tangent there is

x - y + 2 (x + 1) - 3 (y - 1) - 12 = 0

, which simplifies to

3 x - 4 y - 7 = 0

. Any circle touching the given one at that point belongs to the family

(x - 1)^{2} + (y + 1)^{2} + λ (3 x - 4 y - 7) = 0

. Forcing it through

(4, 0)

gives

10 + 5 λ = 0

, so

λ = - 2

, and the circle becomes

x^{2} + y^{2} - 8 x + 10 y + 16 = 0

, whose radius is

16 + 25 - 16 = 25 = 5

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option

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