The equation of the circle having radius 5 and touching the circle $x^2+y^2-2x-4y-20=0$ at $(5,5)$ is

The centre of the given circle is $(1, 2)$ and its radius is 5. Since the radii of the two circles are equal. Therefore, the two circles will touch externally and the point of contact will lie mid-way between the two centres. Let the coordinates of the centre of the required circle be$(h, k)$. Then,

$\frac{h+1}{2}=5$ and $\frac{k+2}{2}=5\Rightarrow h=9$ and $k=8$

Thus, the centre of the required circle is (9, 8). Its equation is

$(x-9{)}^2+(y-8{)}^2=5^2\Rightarrow x^2+y^2-18x-16y+120=0$