Easy to guess, somewhat hard to prove!

$x^{2}$ + $a^{2}$ = 1

$y^{2}$ + $b^{2}$ = 1

$x^{2}$ + $a^{2}$ + $y^{2}$ + $b^{2}$ = 2

$x^{2}$ + $a^{2}$ + $y^{2}$ + $b^{2}$ = 2(( $a\times x$ ) + ( $b\times y$ ))

$x^{2}$ + $a^{2}$ + $y^{2}$ + $b^{2}$ - 2( $a\times x$ ) - 2( $b\times y$ ) = 0

$(x-a)^{2}$ + $(y-b)^{2}$ = 0

⇒ $x = a$ ; $y = b$

therefore,

$x^{2}$ + $a^{2}$ = 1

$a^{2}$ + $a^{2}$ = 1

2 $a^{2}$ = 1

thus, $a^{2}$ = $\frac{1}{2}$

similarly , $b^{2}$ = $\frac{1}{2}$

thus,

$a^{2}$ + $b^{2}$ =1

From Cauchy Schwarz's inequality, we get

$(a\times x + b\times y)^{2} \leq \left(a^{2}+b^{2}\right)\left(x^{2}+y^{2}\right)$

Since $x^{2}+a^{2}+y^{2}+b^{2} = 2$ , we get $x^{2}+y^{2} = 2-\left(a^{2}+b^{2}\right)$

Substitute back into inequality we get

$1 \leq (a^{2}+b^{2})(2-(a^{2}+b^{2}))$

$((a^{2}+b^{2})-1)^{2} \leq 0$

But for any real number $k$ , $k^{2} \geq 0$ .

Therefore, $a^{2}+b^{2}-1 = 0$ is the only case.

Which gives $a^{2}+b^{2} = \boxed{1}$ ~~~

In similar ways, we also get $x^{2}+y^{2} = 1$ .

Using Cauchy-Schwarz inequality:

$x^{ 2 }+a^{ 2 }\ge 2ax\\ { y }^{ 2 }+{ b }^{ 2 }\ge 2by\\ \Rightarrow x^{ 2 }+a^{ 2 }+{ y }^{ 2 }+{ b }^{ 2 }\ge 2ax+2by$

As $ax + by = 1$ , we obtain: $x^{ 2 }+a^{ 2 }+{ y }^{ 2 }+{ b }^{ 2 }\ge 2$

Since ${ x }^{ 2 }+{ a }^{ 2 }={ y }^{ 2 }+b^{ 2 }=1$ , which means ${ x }^{ 2 }+{ a }^{ 2 }+{ y }^{ 2 }+b^{ 2 }=2$ , the equal sign must occur for each inequality mentioned above.

Hence: $x=a, y=b$

$x^2 + a^2 + y^2 + b^2 = 2({a^2 + b^2})=2$

$a^2 + b^2 = 1$