The circle has equation [tex]x^2+y^2=49[/tex]. The slope of the tangent line to the graph of [tex]y(x)[/tex] at some point [tex](x_0,y_0)[/tex] is the derivative [tex]\dfrac{\mathrm dy}{\mathrm dx}[/tex] at that point. We can find this using the chain rule:
[tex]2x+2y\dfrac{\mathrm dy}{\mathrm dx}=0\implies\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac xy[/tex]
so that the tangent line to the circle at [tex](5,2\sqrt6)[/tex] has slope
[tex]\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac5{2\sqrt6}[/tex]
Then the tangent line has equation
[tex]y-2\sqrt6=-\dfrac5{2\sqrt6}(x-5)\implies\boxed{y=-\dfrac{5x}{2\sqrt6}+\dfrac{49}{2\sqrt6}}[/tex]
