Find the real solutions of the equations:$x+y=2$, $xy-z^2=1$

Can anyone help with this? I will show you what I did.

Find the real solutions of the equation:$$x+y=2\tag{1}$$ $$xy-z^2=1 \tag{2}$$

From equation (2), I have $$xy=1+z^2 \tag{3}$$ From equations (1) and (3), I have a new equation $$r^2-2r+(1+z^2)=0$$ The roots of this equation are the solutions. How do I find the root?

Solution 1:

Hint: From $r^2-2r+(1+z^2)=0$, you can have $$(r-1)^2+z^2=0.$$ Hence $r=1$ and $z=0$.

Solution 2:

The problem can receive a straightforward geometric interpretation/solution by considering in the plane the fixed straight line $(L)$ with equation $x+y=2$ and the (parametricaly defined) hyperbola $(H_z)$ defined by $xy=1+z^2$.

Curve $(H_0)$ is tangent to (D) in $A$. It is the unique $(H_z)$ that has a common point with $(L)$. All other curves $(H_z)$ ($z\neq 0$) do not intersect $(L)$. See graphics where four hyperbolas $(H_z)$ for $z=0, z=\pm 0.5, z=\pm 1, z=\pm 1.5$ (red, green, brown, blue curves resp.) are represented.

Thus the solution is $(x,y,z)=(1,1,0)$ where $(x,y)=(1,1)$ are the coordinates of $A$, value $z=0$ coming from the particular $(H_z=H_0).$

Solution 3:

Your quadratic is right: you must have in fact $$r^2-2r+(1+z^2)=0\iff r=1\pm\sqrt{1-(1+z^2)}$$ and because you are interested in real solutions the only possibility is $z=0$

Thus $$r=1\Rightarrow (x,y,z)=(1,1,0)$$