I need to find a point where a circumference intersects an ellipse.
The equation of ellipse is $$\frac{(x-x_{1})^2}{a^2}+\frac{(y-y_{1})^2}{b^2}=1$$ and the equation of circumference is $$(x-x_{0})^2+(y-y_{0})^2=r^2$$ where $(x_{0},y_{0})$ and $(x_{1},y_{1})$ are the centers of the circumference and of the ellipse respectively.
I've been trying to find $(x, y)$ of intersection using these formula's above, but stuck.
If someone already solve this problem Please help.
Further to Arthur's comment, we initially obtain (from the equation of the circle) $$x = x_0 \pm \sqrt{r^2 - (y - y_0)^2}.$$
Substituting this/these value(s) for $x$ in the equation for the ellipse, we get $$\dfrac{\left((x_0 \pm \sqrt{r^2 - (y - y_0)^2}) - x_1\right)^2}{a^2} + \dfrac{(y - y_1)^2}{b^2} = 1.$$
Now, I make the substitutions $$A = x_0$$ $$B = x_1$$ $$C = y_0$$ $$D = y_1$$ so that the last equation above takes the form: $$\dfrac{\left((A \pm \sqrt{r^2 - (y - C)^2}) - B\right)^2}{a^2} + \dfrac{(y - D)^2}{b^2} = 1.$$
At this point, why not ask for help from WolframAlpha? The WolframAlpha computation for $y$ in $$\dfrac{\left((A - \sqrt{r^2 - (y - C)^2}) - B\right)^2}{a^2} + \dfrac{(y - D)^2}{b^2} = 1.$$ is here. The WolframAlpha computation for $y$ in $$\dfrac{\left((A + \sqrt{r^2 - (y - C)^2}) - B\right)^2}{a^2} + \dfrac{(y - D)^2}{b^2} = 1.$$ is here.
Afterwards, substitute the values for $y$ obtained by WolframAlpha into the very first equation above, to get the corresponding values for $x$. You know the drill.