prove algebraically that the straight line with equation x=2y+5 is a tangent to the circle with equation x^2+y^2=5I have gotten 4/5 marks but I need to get 5/5Thank you!

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).