Differentiate both sides of the equation of the circle with respect to [tex]x[/tex], treating [tex]y=y(x)[/tex] as a function of [tex]x[/tex]:
[tex]x^2+y^2=5\implies2x+2y\dfrac{\mathrm dy}{\mathrm dx}=0\implies\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac xy[/tex]
This gives the slope of any line tangent to the circle at the point [tex](x,y)[/tex].
Rewriting the given line in slope-intercept form tells us its slope is
[tex]x=2y+5\implies y=\dfrac12x-\dfrac52\implies\mathrm{slope}=\dfrac12[/tex]
In order for this line to be tangent to the circle, it must intersect the circle at the point [tex](x,y)[/tex] such that
[tex]-\dfrac xy=\dfrac12\implies y=-2x[/tex]
In the equation of the circle, we have
[tex]x^2+(-2x)^2=5x^2=5\implies x=\pm1\implies y=\mp2[/tex]
If [tex]x=-1[/tex], then [tex]-1=2y+5\implies y=-3\neq2[/tex], so we omit this case.
If [tex]x=1[/tex], then [tex]1=2y+5\implies y=-2[/tex], as expected. Therefore [tex]x=2y+5[/tex] is a tangent line to the circle [tex]x^2+y^2=5[/tex] at the point (1, -2).
