A geometry problem by A Former Brilliant Member

$x^2+y^2=m^2$ $\color{#20A900}(1)$

$x+y=1$ $\color{#20A900}(2)$

So that the graphs of $\color{#20A900}(1)$ and $\color{#20A900}(2)$ are tangent to each other, the solutions of the system $[\color{#20A900}(1),\color{#20A900}(2)]$ must be identical.

Substitute $y=1-x$ in $\color{#20A900}(1)$ .

$x^2 + (1-x)^2 = m^2$

$x^2 +1-2x+x^2=m^2$

$2x^2 - 2x + 1 - m^2 = 0$ $\color{#20A900}(3)$

Note that $\color{#20A900}(3)$ must have equal roots. Hence, the discriminant must be zero. $\implies$ $\boxed{b^2-4ac=0}$

$a=2$ , $b=-2$ and $c=1-m^2$

$(-2)^2-4(2)(1-m^2)=0$

$m^2=\frac{1}{2}$

$a+b=1+2=$ $\boxed{\color{#D61F06}\large3}$ $\color{#69047E}answer$