# How do you write an equation for the line tangent to the circle x^2 + y^2 = 29 at the point (2, 5)?

May 11, 2016

$2 x + 5 y = 29$

#### Explanation:

If ${x}^{2} + {y}^{2} = 29$
then by taking the derivative of both sides (to obtain the general slope)
$\textcolor{w h i t e}{\text{XXX}} 2 x + 2 y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
and
the slope is
$\textcolor{w h i t e}{\text{XXX}} m = \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{y}$

Specifically the slope at $\left(x , y\right) = \left(2 , 5\right)$
will be $m = - \frac{2}{5}$
and using the slope point form for the tangent:
$\textcolor{w h i t e}{\text{XXX}} \left(y - 5\right) = - \frac{2}{5} \left(x - 2\right)$

$\textcolor{w h i t e}{\text{XXX}} 5 y - 25 = - 2 x + 4$

$\textcolor{w h i t e}{\text{XXX}} 2 x + 5 y = 29$ (after conversion to standard form)

Here is a graph of the (top half of the) circle equation and the calculated tangent line.
graph{(x^2+y^2-29)(2x+5y-29)=0 [-7.58, 8.224, -0.3, 7.6]}

(note: I am having some problems with this graph display).