A circle whose radius is $5$ and which touches externally the circle $x^2 + y^2 – 2x – 4y – 20 = 0$ at the point $(5, 5)$ intersects in real distinct points the line

 Centre of the given circle is $A (1, 2)$ and its radius is $\sqrt{1+(2{)}^2+20}=5$. 
Point of contact $P$ is $(5, 5)$. Let $B (h, k)$ be the centre of the required circle of radius $5$, then $P$ is the mid-point of $AB$, so that

$\frac{h+1}{2}=5$  and   $\frac{k+2}{2}=5\Rightarrow h=9,k=8$
and an equation of the required circle is
     $x^2+y^2-18x-16y+120=0$
If $x=0, y^2-16y+120=0$ does not give real values of $y$
If $y=0,x^2-18x+120=0$ does not give real values of $x$
If          $y=x,2x^2-34x+120=0$
or   $x^2-17x+60=0 \Rightarrow x=5,12$
This shows that the circle intersects the line y = x at two real distinct points.