Find the co-ordinate of points on the $x-axis$ which are at a distance of $17$units from the point $(11,-8)$

Co-ordiantes of point on $x-$axis be $(x,0)$ which is at a distance of $17$units from the point $(11,-8)$

Distance between the points $(x_1,y_1)$ and $(x_2,y_2)$ is $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

It is given that the distance between points $(x,0)$ and $(11,-8)$ is $17$ units.

$\Rightarrow \sqrt{{(11-x)}^2+{(-8-0)}^2}=17$

$\Rightarrow 121+x^2-22x+64=289$

$\Rightarrow x^2-22x-104=0$

$\Rightarrow x^2-26x+4x-104=0$

$\Rightarrow x(x-26)+4(x-26)=0$

$\therefore x=-4,26$

Hence the two points are $(-4,0)$ and $(26,0)$