Given that a > 2 b > 0 and that the line y = m x - b 1 + m 2 is a common tangent to the circles x 2…

The equation of circles are given as,

$x^2+y^2=b^2$

$(x-a{)}^2+y^2=b^2$

The equation of the comment tangent for $1\text{st}$ is,

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

The equation of the common tangent for $2\text{nd}$ circle is,

$y=m\left(x-a\right)+b\sqrt{1+m^2}$

Equate the common tangents,

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

$2b\sqrt{1+m^2}=am$

$a^2{m}^2=4b^2+4b^2{m}^2$$$

$m^2\left(a^2-4b^2\right)=4b^2$

The value of $m$ is,

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