4 Variables, 3 Equations

$\begin{cases} a^2 + x^2 = 1 & ...(1) \\ b^2 + y^2 = 1 & ...(2) \\ ax + by = 1 & ...(3) \end{cases}$

$\begin{aligned} (1)+(2)+2\times(3): \quad x^2+ 2ax + a^2 + y^2 + 2by + b^2 & = 4 \\ (x+a)^2 + (y+b)^2 & = 4 & ...(4) \end{aligned}$

Equation (4) is a circle with center at $C(-a,-b)$ and a radius of 2.

Similarly,

$\begin{aligned} (1)+(2)-2\times(3): \quad x^2- 2ax + a^2 + y^2 - 2by + b^2 & = 0 \\ (x-a)^2 + (y-b)^2 & = 0 & ...(5) \end{aligned}$

Equation (5) is a point $P(a,b)$ . To satisfy equations (4) and (5), it means $P$ is on the circumference of the circle. That is $(x,y)$ is the point $P(a,b)$ or $x=a$ and $y=b$ . Then from $(3): \ a^2 + b^2 = \boxed 1$ .

Trig solution:

$a^2 + x^2 = 1$ . Choose $\theta$ such that $a = \cos(\theta), x = \sin(\theta)$ . (I.e. $\theta = \pm \arctan(x/a)$ .)

Similarly, choose $\psi$ such that $b = \cos(\psi)$ and $y = \sin(\psi)$ .

Note that $ax = \cos(\theta)\sin(\theta) = \frac{1}{2} \sin(2\theta)$ . Also, $by = \cos(\psi)\sin(\psi) = \frac{1}{2} \sin(2\psi)$

The third equation is thus: $\frac{1}{2} \sin(2\theta) + \frac{1}{2} \sin(2\psi) = 1$ . The only real solutions are when $\sin(2\theta) = \sin(2\psi) = 1$ , i.e. $\theta, \psi \in \{\frac{\pi}{4}, \frac{5\pi}{4}\}$ . For either choice of $\theta$ or $\psi$ , it's true that

$a^2 = ( \pm \frac{\sqrt{2}}{2})^2 = \frac{1}{2}$ , and

$b^2 = ( \pm \frac{\sqrt{2}}{2})^2 = \frac{1}{2}$ .

Thus $a^2 + b^2 = 1$ .

Elementary solution $a^2+x^2=1$ $b^2+y^2=1$ $2ax+2by=2$ add $Eq(1) +Eq(2) -Eq(3)$ you end up with $(a-x)^2+(b-y)^2=0$ this implies that LHS is a purely positive number and the only way it can equal $0$ is when $x=a$ and $b=y$ from there onwards realize that we have 4 variables and 4 equations. meaning a solution can be obtained.