Question

What is the line tangent to the circle at the given point x^2+y^2=13 at (2,3)

Answer 1

well, the derivative is the same, since 13 as well as 20, are both constants, so they simply zero out anyway


`\bf x^2+y^2=13\implies 2x+2y\cfrac{dy}{dx}=0\implies x+y\cfrac{dy}{dx}=0 \\\\\\ \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}}`