Ireland

First of all, the reflection $(x^{\prime },y^{\prime })$ of $(x_0,y_0)$ in $L$ is given by $$x^{\prime }=\frac{(1-m^2)x_0+2m{y}_0}{1+m^2},y^{\prime }=\frac{2m{x}_0+(m^2-1)y_0}{1+m^2}.$$ This is so because $$x^{\prime }+x_0=\frac{2(x_0+m{y}_0)}{1+m^2},y^{\prime }+y_0=\frac{2m(x_0+m{y}_0)}{1+m^2}.$$ Thus the reflection of $(1,1)$ in $L$ has coordinates $$x^{\prime }=\frac{1+2m-m^2}{1+m^2},y^{\prime }=\frac{m^2+2m-1}{1+m^2}.$$ Next, if $m+1\ne 0$, the slope of the line joining $(x^{\prime },y^{\prime })$ to $(-1,1)$ is given by $$\mu =\frac{y^{\prime }-1}{x^{\prime }+1}=\frac{m^2+2m-1-(1+m^2)}{1+2m-m^2+1}=\frac{m-1}{m+1}.$$ Hence, the line joining $(-1,1)$ and $(x^{\prime },y^{\prime })$ has equation $$(m+1)(y-1)=(m-1)(x+1).$$ This line meets $L$ at the point $P$ with coordinates $$x=\frac{2m}{m^2+1},y=\frac{2m^2}{m^2+1}.$$ Since $$x^2+(y-1{)}^2=\frac{4m^2}{(m^2+1{)}^2}+{\left(\frac{m^2-1}{m^2+1}\right)}^2=\frac{4m^2+m^4-2m^2+1}{(m^2+1{)}^2}=1,$$ the point $P$ traverses an arc of the circle of unit radius centred at $(0,1)$.

Second Solution: To solve this problem using synthetic geometry, we denote the unit circle centred at $(0,1)$ by $C$ and let $A=(0,0)$, $B=(1,1)$ and $C=(-1,1)$. The line $L$ intersects $C$ at $A$ and a second point $P$. Let $B^{\prime }$ be the intersection point of the line $PC$ with the line which is perpendicular to $L$ and passes through $B$. Let $M$ be the intersection point of $PA$ and $B{B}^{\prime }$.

The statement we have to show is then equivalent to $|BM|=|M{B}^{\prime }|$. We obtain this as follows. Because the two arcs $CA$ and $AB$ are equal (both are quarter circles or $\pi /2$), we have $\angle CPA=\angle APB$ if $P$ is on the arc $BC$ not containing $A$ and we have $\angle CPA=180^{\circ }-\angle BPA=\angle MPB$ if $P$ is on the arc $AB$ not containing $C$. The case where $P$ is on the arc $AC$ not containing $B$ is similar and if $P$ coincides with $A$, $B$ or $C$ the statement is obvious. Because $\angle PM{B}^{\prime }=\angle BMP$ (both are right angles), the two triangles $BPM$ and $P{B}^{\prime }M$ share the common side $PM$ and have equal adjacent angles, hence are congruent. This shows $|BM|=|M{B}^{\prime }|$.