The circle is $x^2+y^2=25$, so its centre is $(0,0)$ and radius $r=5$. The external point is $P=(7,1)$. First, calculate the length of the tangent $PT$ from $P$ to the circle. $PT^2 = x_1^2+y_1^2-r^2 = 7^2+1^2-25 = 49+1-25 = 25$. So, $PT = 5$. The length of the chord of contact $L$ is given by the formula $L = \frac{2r \cdot PT}{\sqrt{r^2+PT^2}}$. Substitute $r=5$ and $PT=5$: $L = \frac{2 \times 5 \times 5}{\sqrt{5^2+5^2}} = \frac{50}{\sqrt{25+25}} = \frac{50}{\sqrt{50}}$. Simplify $\sqrt{50} = 5\sqrt{2}$. $L = \frac{50}{5\sqrt{2}} = \frac{10}{\sqrt{2}}$. Rationalizing the denominator: $L = \frac{10\sqrt{2}}{2} = 5\sqrt{2}$. Thus, the length of the chord of contact is $5\sqrt{2}$.