How do I find the equation of a tangent and the normal given one point on a circle?

Question

How do I find an equation of the tangent and the normal at the point $(1, -2)$ on the circle $x^2 + y^2 = 5$?

Answer 1

Differentiate the equation w.r.t x, $$\frac{dy}{dx}=\frac{-x}{y}$$ Putting $(1,-2)$, $$\frac{dy}{dx}=\frac{1}{2}$$ Now, you got slope and a point.Apply point slope form, $$(y+2)=0.5(x-1)$$ This is the equation of tangent. $$$$ Since, the tangent is perpendicular to normal; therefore, $$\frac{dy}{dx}=\frac{y}{x}$$ Putting $(1,-2)$, $$\frac{dy}{dx}=\frac{-2}{1}$$ $$\frac{dy}{dx}=-2$$ Apply point slope form, $$(y+2)=-2(x-1)$$ This is the equation of Normal.

Answer 2

The normal line passes through the given point $(1,-2)$ and the centre of the circle $(0,0)$, that is $y=-2x$.

The tangent line passes through the given point $(1,-2)$ and it is orthogonal to the normal line. Hence we obtain $y=\frac{1}{2}(x-1)-2$.

Answer 3

Start with $$ y=-\sqrt{5-x^2}. $$ Then $$ \frac{dy}{dx}=\frac{x}{\sqrt{5-x^2}}. $$ At the point $(1,-2)$, the slope of the tangent line is $$ m_t=\frac{1}{\sqrt{5-1^2}}=1/2. $$ The desired tangent line is $$ y+2=\frac{1}{2}(x-1) $$ which yields $$ x-2y-5=0. $$ For the normal line we have $$ y+2=-2(x-1) $$ which yields $$ 2x+y=0. $$