If a > 2b > 0, then positive value of m for which y = mx - b√(1 + m^2) is a common tangent to x^2 + y^2 = b^2 and (x - a)^2 + y^2 = b^2 is

If a > 2b > 0, then positive value of m for which y = mx -… — JEE Main 2002 Mathematics PYQ

AIEEE 2002

a > 2b > 0

, then positive value of

m

for which

y = mx - b\sqrt{1 + m^2}

is a common tangent to

x^2 + y^2 = b^2

and

(x - a)^2 + y^2 = b^2

A.
$\dfrac{2b}{\sqrt{a^2 - 4b^2}}$
B.
$\dfrac{\sqrt{a^2 - 4b^2}}{2b}$
C.
$\dfrac{2b}{a - 2b}$
D.
$\dfrac{b}{a - 2b}$

Show answer and solution

Correct option: A

Solution

The line y = mx - b√(1+m²) is already tangent to x²+y²=b² (distance from origin = b·√(1+m²)/√(1+m²) = b ✓).

For it to also be tangent to (x-a)²+y²=b², the distance from (a,0) to the line must equal b.

Line: mx - y - b√(1+m²) = 0

Distance = |ma - b√(1+m²)| / √(1+m²) = b

So |ma - b√(1+m²)| = b√(1+m²)

Since a > 2b > 0 and m > 0, take the case ma = 2b√(1+m²):

m²a² = 4b²(1+m²)
m²(a²-4b²) = 4b²
m = 2b/√(a²-4b²)

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