what does it mean if $\text{Cov}(X,Y)$ = $\text{Cov}(Y,Y)$ = $\text{Var}(Y)$ If $\text{Cov}(X,Y)$ = $\text{Cov}(Y,Y)$ = $\text{Var}(Y)$

then what can be said about X and Y?

My rusty math skills took me this far:

$$\text{Cov}(X,Y) = E[(X-\mu_X)(Y-\mu_Y)]$$

so we get 

$$E[(X-\mu_X)(Y-\mu_Y)] = E[(Y-\mu_Y)(Y-\mu_Y)]$$

Does this mean that $(X-\mu_X) = (Y-\mu_Y)$ ? 

How should I interpret that?

Intuitive answers are also more than welcome.
We know that covariance is a bilinear function. So
 $$Cov(X,Y)=Cov(Y,Y)$$
$$\Longleftrightarrow Cov(X,Y)-Cov(Y,Y)=0$$
$$\Longleftrightarrow Cov(X-Y,Y)=0$$

Therefore, $X-Y$ and $Y$ are uncorrelated.

EDIT
Dividing both sides by the product of standard deviations of $X$ and $Y$, we get
$$Cor(X,Y)=\frac{SD(Y)}{SD(X)}$$

Therefore, we observe two more things:

1) X and Y are positively correlated.

2)$SD(Y)&lt;SD(X)$

Hope it helps:)
Obviously, X and Y are not independent. So $(X-\mu_X) = (Y-\mu_Y)$ is uncorrect.