Let intersection points of these two parabolas are $$\mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_1\right) \& \mathrm{~B}\left(\mathrm{x}_2, \mathrm{y}_2\right)$$

$\because$ equation of parabola I and II are given below

$$\begin{aligned} & \therefore(\mathrm{x}-4)^2+(\mathrm{y}-3)^2=\mathrm{x}^2 \quad\text{..... (1)}\\ & \&(\mathrm{x}-4)^2+(\mathrm{y}-3)^2=\mathrm{y}^2 \quad\text{..... (2)} \end{aligned}$$

Here $A\left(x_1, y_1\right) \& B\left(x_2, y_2\right)$ will satisfy the equation

Also from equations (1) & (2), we get $x=y$ .......(3)

Put $x=y$ in equation (1)

We get $x^2-14 x+25=0$

$$\begin{aligned} & x_1+x_2=14 \\ & x_1 x_2=25 \\ & \therefore A B^2=\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2 \\ & =2\left(x_1-x_2\right)^2 \\ & =2\left[\left(x_1+x_2\right)^2-4 x_1 x_2\right] \\ & =192 \end{aligned}$$