# Implicit differentiation

## Tangent lines of a circle

- Recall that the equation of the circle \(C\) with radius 5 and centre \(O(0,0)\) is \[
x^2+y^2=25.
\]
- Let us find equations of its tangent lines.
- One way would be to separate the circle into the two graphs \(y=\pm\sqrt{25-x^2}\), and derivate the functions.
*Implicit differentiation* gives an easier way: you can derivate the equation of the circle itself with respect to \(x\).
- What you need to look out for is that when derivating an expression containing \(y\), you need to derivate that as \(y(x)\) and thus use the chain rule.
- Let's do this! We get \[
2x+2yy'=0.
\]
- Solving for \(y'\) gives \(y'=-\frac{x}{y}\). This is saying that the slope of the tangent line line at \(P(a,b)\) of \(C\) is \(y'(x=a)=-\frac{a}{b}\).
- For example, the tangent line at \(P(3,4)\) has equation \(y-4=-\frac34(x-3)\).
- Exercises. 3.5: 2, 4, 10, 12, 26, 34, 46