Question

Find an equation of the tangent line to the circle x squared plus y squared equals 49 at the point
`(5,2√(6))`?

Answer 1

The circle has equation
`x^2+y^2=49`. The slope of the tangent line to the graph of
`y(x)` at some point
`(x_0,y_0)` is the derivative
`(\mathrm dy)/(\mathrm dx)` at that point. We can find this using the chain rule:

`2x+2y(\mathrm dy)/(\mathrm dx)=0\implies(\mathrm dy)/(\mathrm dx)=-\frac xy`

so that the tangent line to the circle at
`(5,2\sqrt6)` has slope

`(\mathrm dy)/(\mathrm dx)=-\frac5{2\sqrt6}`

Then the tangent line has equation

`y-2\sqrt6=-\frac5{2\sqrt6}(x-5)\implies\boxed{y=-(5x)/(2\sqrt6)+(49)/(2\sqrt6)}`