1. Find the circle $C$

Since the circle has radius $2$, lies in the second quadrant, and touches both coordinate axes, its center must be at $(-2,2).$ So circle $C$ has center $O(-2,2)$ and radius $R=2$.

2. Given second circle

The second circle has center $P(2,5)$ and radius $r$.

We first find the distance between the centers: $OP=\sqrt{(2-(-2))^2+(5-2)^2}=\sqrt{4^2+3^2}=\sqrt{25}=5.$

So the distance between centers is $d=5.$

3. Condition for two circles to intersect at exactly two points

For two circles with radii $R$ and $r$, and center distance $d$, they intersect at exactly two points when $|R-r|<d<R+r.$

Here, $R=2$, $d=5$. So:

$|2-r|<5 \quad \text{and} \quad 5<2+r.$

4. Solve the inequalities

(i) From $5<2+r$

$r>3.$

(ii) From $|2-r|<5$

This gives $-5<2-r<5.$ Subtract $2$: $-7<-r<3.$ Multiply by $-1$ and reverse signs: $-3<r<7.$

Combining with $r>3$, we get $3<r<7.$

Hence, the set of all possible values of $r$ is $(\alpha,\beta)=(3,7).$

So, $3\beta-2\alpha=3(7)-2(3)=21-6=15.$

5. Check options

The value is $15,$ which corresponds to Option D.

6. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.