The length of the tangent drawn from any point on the circle x^ 2 +y^ 2 +2 f y+λ=0 to the circle ^ 2 + ^ 2 +2 fy+μ=0, where μ > λ > 0, is

Correct answer: A. μ-λ Let the radius of the first circle be CT=r_ 1 . Also, let the radius of the second circle be CP=r _ 2 In the triangle PCT, T is a right angle aligned So, P T &= P C^ 2 -C T^ 2 = r_ 1 ^ 2 -r_ 2 ^ 2 &= (f^ 2 -λ)-(f^ 2 -μ) = μ-λ aligned

BITSAT 2012

Exam: BITSAT
Year: 2012
Subject: Mathematics
Chapter: Circle

BITSAT 2012 · Mathematics · Circle

The length of the tangent drawn from any point on the circle x^ 2...

The length of the tangent drawn from any point on the circle

x^{2}+y^{2}+2 f y+\lambda=0

to the circle

\mathrmx^{2}+\mathrmy^{2}+2 \mathrm{fy}+\mu=0,

where

\mu > \lambda > 0,

The length of the tangent drawn from any point on the circle x^ 2 +y^ 2 +2 f y+λ=0 to the circle ^ 2 + ^ 2 +2 fy+μ=0,

Options

A $\sqrt{\mu-\lambda}$
B $\sqrt{\mu+\lambda}$
C $\sqrt{\mu^{2}-\lambda^{2}}$
D $\mu+\lambda$

Correct answer

A. $\sqrt{\mu-\lambda}$

Step-by-step solution

Let the radius of the first circle be

\mathrm{CT}=\mathrm{r}_{1}. Also, let

the

radius of the second circle be

\mathrm{CP}=\mathrm{r}

_{2}

In the triangle

\mathrm{PCT}, \mathrm{T} is a right angle <img src="/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fbitsat%252FVx6mkKDSQTkMXkiFF9Z4ygG5UGK_UM5zCD04GOUxX60.original.fullsize.png%3Falt%3Dmedia&clean=1&v=wm2" data-qx-orig-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fbitsat%2FVx6mkKDSQTkMXkiFF9Z4ygG5UGK_UM5zCD04GOUxX60.original.fullsize.png?alt=media" data-qx-storage-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fbitsat%2FVx6mkKDSQTkMXkiFF9Z4ygG5UGK_UM5zCD04GOUxX60.original.fullsize.png?alt=media"><br>

\begin{aligned} \text{ So, } P T &=\sqrt{P C^{2}-C T^{2}}=\sqrt{r_{1}^{2}-r_{2}^{2}} \\ &=\sqrt{\left(f^{2}-\lambda\right)-\left(f^{2}-\mu\right)}=\sqrt{\mu-\lambda} \end{aligned}

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