Question

Can any one ANS this question........of Circle ⭕

Answer 1

answer : option (1)

explanation : $y=mx-b\sqrt{1+m^2}$ is common tangent to x² + y² = b² and (x - a)² + y² = b² .

here you should notice that, radius of both circle is b.

so, distance between centre of circle and common tangent will be b for both circles.

e.g., for circle : x² + y² = b² centre is (0,0)

b = $\frac{|0-m.0+b\sqrt{1+m^2}|}{\sqrt{-m)^2+1^2}}$

= $\frac{|b\sqrt{1+m^2}|}{\sqrt{m^2+1}}$.... (1)

similarly, for circle : (x - a)² + y² = b² centre is (a,0)

b = $\frac{|0-ma+b\sqrt{1+m^2}|}{\sqrt{(-m)^2+1^2}}$

= $\frac{|-ma+b\sqrt{1+m^2}|}{\sqrt{m^2+1}}$.....(2)

from equations (1) and (2),

$|b\sqrt{1+m^2}|=|-ma+b\sqrt{1+m^2}|$

or, $b\sqrt{1+m^2}=ma-b\sqrt{1+m^2}$ [ we didn't choose m = 0, because question is demanding positive value of m]

or, $2b\sqrt{1+m^2}=ma$

squaring both sides,

or, $4b^2+4b^2m^2=m^2a^2$

so, $m=\frac{2b}{\sqrt{a^2-4b^2}}$

hence, option (1) is correct.