How to solve and oval and a circle

Question

Given an oval $x^2+\left(1.4^x 1.6y\right)^2=1$ and a circle $x^2+y^2=0.5$. I want to find the points where they intersect. I am trying to solve $x^2+\left(1.4^x 1.6y\right)^2-1=x^2+y^2-0.5$ in Mathematica but I am getting $y$ in terms of x. But what is the general way to solve these simultaneous equations and get these 4 intersection points as shown in the figure below ?

Answer 1

you Can eliminate $y$ by $$y=\pm\sqrt{\frac{1}{2}-x^2}$$ try it! for your Control i got this here $$x \approx -0.09053709354, y \approx -.7012866994$$

Answer 2

I do not think that analytical solutions could be obtained and numerical method should be used.

Considering the more general case of $$x^2+ \left(a^x \,by \right)^2=1\implies x^2+a^{2x}b^2y^2=1\tag 1$$ $$x^2+y^2=r^2 \tag 2$$ eliminate $y^2$ from $(1)$ to get $$b^2 a^{2 x} \left(r^2-x^2\right)+x^2=1\tag 3$$ So, search for the zero of function $$f(x)=b^2 a^{2 x} \left(r^2-x^2\right)+x^2-1\tag 4$$ Graphing will show more or less the roots and Newton Raphson will polish them.

Considering your case $a=1.4$, $b=1.6$, $r^2=0.5$ a plot will show you three roots (around $-1.75$, $-0.25$ and $0.5$). The first one must be discarded. For the two other, Newton iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & -0.2500000000 \\ 1 & -0.2574440860 \\ 2 & -0.2574344423 \end{array} \right)$$

$$\left( \begin{array}{cc} n & x_n \\ 0 & 0.5000000000 \\ 1 & 0.5736985982 \\ 2 & 0.5637257950 \\ 3 & 0.5635181657 \end{array} \right)$$ Once $x$ is known, use $(2)$ to get the corresponding $y$'s.

If you use $b \leq 1.3453$, the plot of the function will show no acceptable root; then, no intersection.

Edit

If you do not change the values of parameters $a=1.4$ and $b=1.6$ I found here $$x^2+a^{2x}b^2y^2=1$$ and you want to know for which value of $r$ there will be intersection with $$x^2+y^2=r^2$$ you need to consider solving for $x$ and $r$ the two equations $$f(x)=b^2 a^{2 x} \left(r^2-x^2\right)+x^2-1=0\tag 5$$ $$f'(x)=2 b^2 a^{2 x} \log (a) \left(x^2-r^2\right)+2 b^2 x a^{2 x}+2 x=0\tag 6$$ From $(6)$ we can eliminate $r^2$ and plug its expression in $(5)$ to get a single equation in $x$ which, using whole numbers write $$x^2+\frac{\left(64 \left(\frac{49}{25}\right)^x-25\right) x}{25 \log \left(\frac{7}{5}\right)}-1\tag 7$$ The only acceptable solution is $x\approx 0.173790$ to which corresponds $r\approx0.605985$.

So, if $0.606 < r < 1$, the two curves intersect at four points.