well, the derivative is the same, since 13 as well as 20, are both constants, so they simply zero out anyway
[tex]\bf x^2+y^2=13\implies 2x+2y\cfrac{dy}{dx}=0\implies x+y\cfrac{dy}{dx}=0
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\cfrac{dy}{dx}=\cfrac{-x}{y}\\\\
-------------------------------\\\\
y'(2,3)=\cfrac{-2}{3}\implies y'(2,3)=-\cfrac{2}{3}\\\\
-------------------------------\\\\
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-3=-\cfrac{2}{3}(x-2)
\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}
\\\\\\
y-3=-\cfrac{2}{3}x+\cfrac{4}{3}\implies \boxed{y=-\cfrac{2}{3}x+\cfrac{13}{3}}[/tex]
