A circle C touches the line x = 2y at the point (2,1) and intersects the circle ${\mathrm{C}}_1 : {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{y}-5=0$ at two points P and Q such that PQ is a diameter of C₁. Then the diameter of C is

Circle C touches x = 2y at (2, 1), then equation of circle C is given by

     $\begin{array}{l}(\mathrm{x}-2{)}^2+(\mathrm{y}-1{)}^2+\mathrm{\lambda }(\mathrm{x}-2\mathrm{y})=0\\ \Rightarrow {\mathrm{x}}^2-4\mathrm{x}+4+{\mathrm{y}}^2-2\mathrm{y}+1+\mathrm{\lambda x}-2\mathrm{\lambda y}=0\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+(\mathrm{\lambda }-4)\mathrm{x}+(-2-2\mathrm{\lambda })\mathrm{y}+5=0 ...\left(\mathrm{i}\right)\end{array}$

Since circles C and C₁ intersect each other at PQ, so PQ is common chord.

$\therefore$Equation of PQ will be $\mathrm{C}-{\mathrm{C}}_1=0$

$\Rightarrow \mathrm{x}(\mathrm{\lambda }-4)+\mathrm{y}(-4-2\mathrm{\lambda })+10=0$

Again PQ is diameter of circle C₁.

So, PQ passes through centre of C₁.

Now, equation of C₁ is ${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{y}-5=0$

$\therefore$ Centre of ${\mathrm{C}}_1\text{ is }(0,-1)$

So, PQ passes through $(0,-1)$

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0(\mathrm{\lambda }-4)-1(-4-2\mathrm{\lambda })+10=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4+2\mathrm{\lambda }+10=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{\lambda }=-7\end{array}$

On putting $\mathrm{\lambda }=-7$ in Eq. (i), we get equation of circle C as

$\begin{array}{l}{\mathrm{x}}^2+{\mathrm{y}}^2-11\mathrm{x}+12\mathrm{y}+5=0\\ \therefore \text{ Radius of }\mathrm{C}=\sqrt{{\left(\frac{11}{2}\right)}^2+(-6{)}^2-5}\\ =\sqrt{\frac{12}{4}+36-5}=\sqrt{\frac{245}{4}}=\frac{\sqrt{245}}{2}\\ \therefore \text{ Diameter of }\mathrm{C}=2\times \text{ Radius }=\sqrt{245}=7\sqrt{5}\end{array}$