Consider this family of equations:

$Q={\left(x-7\right)}^4+k{\left(x-7\right)}^2{\left(y-10\right)}^2+{\left(y-10\right)}^4$

The equation is symmetrically centered at the point $(\text{}7\text{},\text{}10\text{})$.

Replace all $x-7$ with $X$ and $y-10$ with $Y$ to shift this center to origin:

$Q=X^4+k{X}^2{Y}^2+Y^4$

Substitute $X=r\cos \left(\theta \right)$ and $Y=r\sin \left(\theta \right)$:

$Q=r^4\left({\cos }^4\left(\theta \right)+k{\cos }^2\left(\theta \right){\sin }^2\left(\theta \right)+{\sin }^4\left(\theta \right)\right)$

$Q=r^4\left(1+\left(k-2\right){\cos }^2\left(\theta \right){\sin }^2\left(\theta \right)\right)$



$0\le {\cos }^2\left(\theta \right){\sin }^2\left(\theta \right)\le \frac{1}{4}$

The minimum of ${\cos }^2\left(\theta \right){\sin }^2\left(\theta \right)$ occurs on axes ($x=0$ or $y=0$). The maximum occurs on diagonals ($y=\pm x$).

We have $k=4$. Therefore, $Q=r^4\left(2{\sin }^2\left(\theta \right){\cos }^2\left(\theta \right)+1\right)$. Since $Q$ values are higher along the diagonals than along the axes, the contour plot will look like closed concentric loops that are stretched along the coordinate axes.

Note that this pattern will be centered at $(\text{}7\text{},\text{}10\text{})$.