1. Since the equation $ax^2-2bx+15=0$ has a repeated root $\alpha$, its discriminant must be zero.

So, $(-2b)^2-4(a)(15)=0$ $4b^2-60a=0$ $b^2=15a$

2. For a quadratic with repeated root, the root is $\alpha=\frac{-(-2b)}{2a}=\frac{2b}{2a}=\frac{b}{a}$

Using $b^2=15a$, we get $a=\frac{b^2}{15}$ Thus, $\alpha=\frac{b}{a}=\frac{b}{b^2/15}=\frac{15}{b}$

3. Also, since $\alpha$ is a root of $ax^2-2bx+15=0,$ substitute $x=\alpha$: $a\alpha^2-2b\alpha+15=0$ But because it is a repeated root, another standard relation is: $\alpha^2=\frac{15}{a}$ Now using $a=\frac{b^2}{15}$, $\alpha^2=\frac{15}{b^2/15}=\frac{225}{b^2}$

Also from $\alpha=\frac{15}{b}$, this matches.

4. Now consider the equation $x^2-2bx+21=0$ whose roots are $\alpha$ and $\beta$.

By Vieta's formulas, $\alpha+\beta=2b, \qquad \alpha\beta=21$

Since $\alpha$ is one root and from above $\alpha=\frac{15}{b}$, use product relation: $\alpha\beta=21$ $\frac{15}{b}\beta=21$ $\beta=\frac{21b}{15}=\frac{7b}{5}$

Now use sum relation: $\alpha+\beta=2b$ $\frac{15}{b}+\frac{7b}{5}=2b$ Multiply by $5b$: $75+7b^2=10b^2$ $3b^2=75$ $b^2=25$

Hence, $\alpha=\frac{15}{b}, \qquad \alpha^2=\frac{225}{25}=9$

5. Since $\alpha\beta=21$, $\beta^2=\frac{21^2}{\alpha^2}=\frac{441}{9}=49$

Therefore, $\alpha^2+\beta^2=9+49=58$

6. Hence the correct option is $\boxed{58}$ which is option $\boxed{B}$.