jmc

A line passing through $(1,3)$ has the form $$y-3=m(x-1),$$Then $x-1=\frac{y-3}{m},$ so $x=\frac{y-3}{m}+1=\frac{y+m-3}{m}.$ Substituting into $y^2=4x,$ we get $$y^2=4\cdot \frac{y+m-3}{m}.$$We can write this as $m{y}^2-4y+(-4m+12)=0.$ Since we have a tangent, this quadratic will have a double root, meaning that its discriminant is 0. Hence, $$16-4(m)(-4m+12)=0.$$This simplifies to $m^2-3m+1=0.$ Let the roots be $m_1$ and $m_2.$ Then by Vieta's formulas, $m_1+m_2=3$ and $m_1{m}_2=1,$ so $$(m_1-m_2{)}^2=(m_1+m_2{)}^2-4m_1{m}_2=9-4=5.$$We know that $y$ is a double root of $m{y}^2-4y+(-4m+12)=0,$ so by completing the square we can see that the corresponding values of $y$ are $y_1=\frac{2}{m_1}=2m_2$ and $y_2=\frac{2}{m_2}=2m_1.$ Then $$x_1=\frac{y_1^2}{4}=m_2^2$$and $$x_2=\frac{y_2^2}{4}=m_1^2.$$Therefore, $A$ and $B$ are $(m_1^2,2m_1)$ and $(m_2^2,2m_2),$ in some order.

So if $d=AB,$ then $$\begin{array}{rl}{d}^2& =(m_2^2-m_1^2{)}^2+(2m_2-2m_1{)}^2\\ & =(m_1+m_2{)}^2(m_1-m_2{)}^2+4(m_1-m_2{)}^2\\ & =3^2\cdot 5+4\cdot 5=65,\end{array}$$so $d=\boxed{\sqrt{65}}.$