Continuum between arithmetic mean, AGM, and geometric mean NOTE: I am aware of this possible duplicate, but my question is slightly different as it also involves arithmetic-geometric mean.
The arithmetic mean  of two numbers is defined as:
$$\text{am}(a,b) = \frac{a}{2}+\frac{b}{2}$$
For example, $\text{am}(3,12) = 7.5$. Similarly, geometric mean of two numbers is defined as:
$$\text{gm}(a,b) = \sqrt{ab}$$
So $\text{gm}(3,12)=6$. The arithmetic-geometric mean is a mean "between" these two, calculated as follows:
$$
a_{0}=a,g_{0}=b\\
a_{n+1}=\text{am}(a_{n},g_{n})\\
g_{n+1}=\text{gm}(a_{n},g_{n})\\
\text{agm}(a,b)=\text{lim}_{n\rightarrow\infty}\;\;(a_n)
$$
For instance, $\text{agm}(3,12) \approx 8.7407$.
Note that $\text{gm}(a,b)\leq \text{agm}(a,b)\leq \text{am}(a,b)$.
My question: I need to define a general mean $M(a,b,c)$ for $0\leq c \leq 1$ that satifies these properties:
$$
M(a,b,0) = \text{am}(a,b)\\
M(a,b,0.5) = \text{agm}(a,b)\\
M(a,b,1) = \text{gm}(a,b)
$$
Power mean is not what I'm looking for, as $PM_{0.5}\;(a,b) \neq \text{agm}(a,b)$ (e.g. $PM_{0.5}\;(3,12) = 8.7464... \neq 8.7407...$).
 A: Standard quadratic interpolation, as suggested by  @Martin-R
\begin{align} 
M(a,b,t)&=
(2t-1)(t-1)\operatorname{AM}(a,b)
\\
&-4t(t-1)\operatorname{AGM}(a,b)
\\
&+t(2t-1)\operatorname{GM}(a,b)
.
\end{align}
Or in expanded form,
\begin{align} 
M(a,b,t)&=
 (\sqrt a+\sqrt b)^2\,t^2
 -(\tfrac32\,a+\tfrac32\,b+\sqrt{ab})\,t
 +\tfrac12\,a+\tfrac12\,b
 \\
 &+\frac{\pi t(1-t)}
 {\displaystyle\int_0^1 \left(\sqrt{1-x^2}\sqrt{(a+b)^2-(a-b)^2\,x^2}\right)^{-1} \, dx }
.
\end{align}

Edit
At the second thought, I must admit that
such interpolation is a bad choice for general $a,b\ge0$.
It's ok for just switching between the three values $t=0,\tfrac12,1$ and works
relatively reasonable for some ranges of $a,b$, but it could
as bad as returning negative output, for example,
$M(1000000,1,0.9)\approx-2081$, so it needs a serious treatment.
A: The geometric mean can be turned to an arithmetic one by a nonlinear mapping,
$$\sqrt{ab}=f^{-1}\left(\frac{f(a)+f(b)}2\right)$$ where $f(x):=\log(x)$ is the logarithm.
Hence a way to construct "intermediate" means is by using a function intermediate between the identity and the logarithm. You could obtain yours by solving the functional equation
$$2f(\text{agm}(a,b))=f(a)+f(b).$$
By homogeneity, it should be enough to solve
$$2f(\text{agm}(a,1))=f(a)+f(1).$$
Of course solving this equation seems arduous. I guess that you can solve it numerically and tabulate $f$. Then you'll have a generalization of the arithmetico-geometric mean for $n$ arguments.
Whether you can find a natural "interpolation" between these three means seems even less tractable.
