Slopes of tangents through $\left(7, 1\right)$ to the circle $x^2+y^2=25$ satisfy the equation 

Let m  be  slope of tangent

Now equation of tangent  passes through (7, 1) is 

$$\begin{gathered} \Rightarrow (\mathrm{y}-1)=\mathrm{m}(\mathrm{x}-7) \\ \Rightarrow \mathrm{mx}-\mathrm{y}+1-7\mathrm{m}=0\to \left(1\right) \\ Given circle x^2+y^2=25 \\ \mathrm{centre}\left(\mathrm{C}\right)=(0, 0) , \mathrm{r}=5 \\ Since \left(1\right) is \tan gent then \\ r=d \\ \Rightarrow r^2=d^2 \\ \Rightarrow 25=\frac{{\left|1-7\mathrm{m}\right|}^2}{1+{\mathrm{m}}^2} \\ \Rightarrow 25+25{\mathrm{m}}^2=1+49{\mathrm{m}}^2-14\mathrm{m} \\ \Rightarrow 24{\mathrm{m}}^2-14\mathrm{m}-24=0 \\ \Rightarrow 12{\mathrm{m}}^2-7\mathrm{m}-12=0 \end{gathered}$$