Given circle $C_1$

$x^2+y^2+2y-5=0$

Complete the square in $y$:

$x^2+(y+1)^2=6$

So, the center and radius of $C_1$ are:

$O_1=(0,-1), \qquad r_1=\sqrt{6}$

Condition that $PQ$ is a diameter of $C_1$

If $P,Q$ are endpoints of a diameter of $C_1$, then the line $PQ$ must pass through the center $O_1=(0,-1)$.

Since $P,Q$ are also the intersection points of circles $C$ and $C_1$, the common chord $PQ$ is the radical axis of the two circles.

Therefore, the radical axis of $C$ and $C_1$ passes through $(0,-1)$.

Use tangent condition for circle $C$

Circle $C$ touches the line $x=2y$ at $(2,1)$.

The line is:

$x-2y=0$

Its slope is $\frac12$, so the radius at the point of contact is perpendicular to this line. A normal vector to the line is $(1,-2)$.

Hence the center of $C$ lies on the line through $(2,1)$ in direction $(1,-2)$:

$\text{Center of }C = (2,1)+\lambda(1,-2)=(2+\lambda,1-2\lambda)$

Let the center be

$O=(h,k)=(2+\lambda,1-2\lambda).$

The radius of $C$ is distance from $O$ to $(2,1)$:

$r=\sqrt{\lambda^2+(-2\lambda)^2}=\sqrt{5}\,|\lambda|$

so

$r^2=5\lambda^2.$

Equation of circle $C$

$x^2+y^2+2gx+2fy+c=0$

with center $(-g,-f)=(h,k)$, so

$g=-h=-(2+\lambda), \qquad f=-k=-(1-2\lambda)=-1+2\lambda.$

Also,

$c=h^2+k^2-r^2.$

Compute:

$h^2+k^2=(2+\lambda)^2+(1-2\lambda)^2$ $=4+4\lambda+\lambda^2+1-4\lambda+4\lambda^2=5+5\lambda^2$

Therefore,

$c=(5+5\lambda^2)-5\lambda^2=5.$

So circle $C$ is

$x^2+y^2-2(2+\lambda)x+2(-1+2\lambda)y+5=0.$

Radical axis of $C$ and $C_1$

Subtract equations:

$\bigl(x^2+y^2-2(2+\lambda)x+2(-1+2\lambda)y+5\bigr)-\bigl(x^2+y^2+2y-5\bigr)=0$

$-2(2+\lambda)x + \bigl(2(-1+2\lambda)-2\bigr)y +10=0$

$-2(2+\lambda)x +(4\lambda-4)y+10=0.$

Since this radical axis passes through $(0,-1)$,

$-2(2+\lambda)(0)+(4\lambda-4)(-1)+10=0$

$-4\lambda+4+10=0$

$-4\lambda+14=0$

$\lambda=\frac{7}{2}.$

Radius and diameter of circle $C$

$r=\sqrt{5}\left|\lambda\right|=\sqrt{5}\cdot \frac72=\frac{7\sqrt{5}}{2}$

Hence diameter is

$2r=7\sqrt{5}.$

Check options

The diameter of $C$ is

$\boxed{7\sqrt{5}}$

This matches Option A.