There are two points $A$ & $B$ which are at a distance of $10$ units from $P$ and farthest ( $10$ unit distance) from the line $4x+3y=17.$

The slope of $AB=\frac{3}{4}$ {Because the slope of $4x+3y=17$ is $-\frac{4}{3}$ }
Let line $AB$ makes an angle $\theta$ with $x$-axis then $\tan \theta =\frac{3}{4}\Rightarrow \left(\cos \theta ,\sin \theta \right)=\left(\frac{\mathrm{}4}{5},\frac{3}{5}\right)$
$\Rightarrow A$ or $B=\left(10\cos \theta -\mathrm{1,10}\sin \theta +7\right)$ or $\Rightarrow A$ or $B=\left(-10\cos \theta -\mathrm{1,}-10\sin \theta +7\right)$
$\Rightarrow A$ or $B=\left(\mathrm{7,13}\right)$ or $\left(-\mathrm{9,1}\right)$