The chord of contact of a point $A\left(x_A,y_A\right)$ of $y^2=4x$ passes through $\left(3,1\right)$, and the point $A$ lies on $x^2+y^2=5^2$ then 

The chord of contact of the point $\left(x_A,y_A\right)$ to the parabola $y^2=4x$ is $S_1=0$
$$\begin{gathered} y{y}_A-2\left(x+x_A\right)=0 \\ 2x-y{y}_A+2x_A=0 \end{gathered}$$

Since it is passing through the point $\left(3,1\right)$

it implies that $6-y_A+2x_A=0$

From the above equaiton $y_A=2x_A+6$

$$\begin{gathered} {\left(x_A\right)}^2+4{x_A}^2+36+24x_A=25 \\ 5{x_A}^2+24x_A+11=0 \end{gathered}$$

Similarly the above equation in terms of $y_A$ is $5{y_A}^2-12y_A-64=0$