If a circle passes through (0, 0), (4, 0) and touches the circle x² + y² = 36 then

$\mathrm{Equation} \mathrm{of} \mathrm{the} \mathrm{circle} \mathrm{passing} \mathrm{through} \left(0,0\right)\mathrm{and}(4,0) \mathrm{is}$

$x\left(x-4\right)+y^2+\lambda y=0\Rightarrow x^2+y^2-4x+\lambda y=0$

This  circle touches $x^2 + y^2= 36$

$C_1{C}_2=r_1-r_2\Rightarrow \sqrt{4+\frac{{\lambda }^2}{4}}=6-\sqrt{4+\frac{{\lambda }^2}{4}}$

$4\left(4+\frac{{\lambda }^2}{4}\right)=36\Rightarrow {\lambda }^2=20$

$x^2+y^2-4x\pm 2\sqrt{5}y=0$

$\therefore \mathrm{c}(2,\sqrt{5})\left(\text{ or }\right)\mathrm{c}(2,-\sqrt{5})$
Length of common chord = distance between (0,0) and (4,0)=4
$\cos \mathrm{\theta }=\frac{20-9-9}{2\times 3\times 3}=\frac{1}{9}$