Let $\tau_{\omega}$ be the inversion with respect to circle $\omega$ and $\omega$ is the unit circle in $\mathbb{E}^2$ with centre $O=(0,0)$. There are three cases:
a. Find $\tau_{\omega}(\ell)$ where $\ell$ is the line $x+y=1$;
b. Find $\tau_{\omega}(\ell)$ where $\ell$ is the line $x=\frac{1}{2}$;
c. Find $\tau_{\omega}(\omega')$ where $\omega'$ is the circumcircle of the triangle $OAB$ with $A=(-1,0)$ and $B=(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$.
I am stuck on a. I know $\ell$ is mapped to a circle $\omega'$ with centre $O'$ such that $\ell$ is the radical axis of $\omega$ and $\omega'$. Since $OO'$ must be orthogonal to the line $x+y=1$, $O'$ must lie on the line $y=x$. However, I have no idea on how to determine the radius of $\omega'$: The line $y=x$ bisects the line $x+y=1$ that connects $(1,0)$ and $(0,1)$ and is orthogonal to $x+y=1$, it seems that the radius can be any value.
Is my idea wrong?
As you know
that the center of your circle lies on line $y=x$ and
that the image of a straight line not passing through the origin is a circle passing through the origin,
you just need to have the image of a point. Take $A(1/2,1/2)$ with $OA=\frac12 \sqrt{2}$ ; its image is $A'$ such that $OA.OA'=1$ and $O,A,A'$ aligned, therefore such that $OA'=\sqrt{2}$, which means $A'=(1,1)$.
Therefore, it is the circle with center $A(1/2,1/2)$ and radius $\frac12 \sqrt{2}$.
Sanity check : the image of point $B(1,0)$ is point $(1,0)$ itself (indeed this point belongs to the inversion circle, which is the set of invariant points). Same thing for point $C(0,1)$.
Remark : please note that this Geogebra-generated figure directly uses inversion under the name "reflection of an "object" with respect to a circle" ; for example "f'=reflect(f,c)" means that the reflection of "object straight line" f=BC with respect to inversion circle c is "object-circle" f'.