hello, we have $(x-1)^2+(y-2)^2\geq 2$ and the equal sign holds if $x=0,y=1$ Sonnhard.

Dr Sonnhard Graubner wrote: hello, we have $(x-1)^2+(y-2)^2\geq 2$ and the equal sign holds if $x=0,y=1$ Sonnhard. Hello Dr Sonnhard Graubner Sir! Note that the equality sign holds true even for $x=2,y=1$

SMOJ wrote: Let $x,y$ be real numbers such that $y=|x-1|$. What is the smallest value of $(x-1)^2+(y-2)^2$? Minimum of $(x-1)^2+(y-2)^2 $ is square of distance from point $(1,2) $ to any of line $y=x-1$ or $y=1-x$ $\implies{(x-1)^2+(y-2)^2 \geq{2}}$ ,equality when $x=0,y=2$ or $x=2,y=1$

this is the case when the circle centered at (1,2) touches the lines y = |x - 1| and symmetrically both the parts, on either side of the line x = 1 have equal area. Thus it is the distance of the point (1,2) from any of the line y = x - 1 or y = 1 - x as mentioned.

Equality holds when $(x,y)=(0,1)$ and $(2,1)$. However, through the distance formula, the actual answer is $2$. Note that the question is asking for the minimum value, not when the equality holds.

the smallest possible answer that I think is 2