# Consider the circle of radius 5 centered at (0,0), how do you find an equation of the line tangent to the circle at the point (3,4)?

Apr 20, 2015

Using calculus:

the equation of the circle is ${x}^{2} + {y}^{2} = 25$

At the point $\left(3 , 4\right)$, we are on the upper semicircle, whose equation is: $y = \sqrt{25 - {x}^{2}}$

So $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{25 - {x}^{2}}} \left(- 2 x\right) = \frac{- x}{\sqrt{25 - {x}^{2}}}$

At $\left(3 , 4\right)$, we get $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 3}{4}$

So the slope of the line is $- \frac{3}{4}$ and the line contains the point $\left(3 , 4\right)$.

$y = - \frac{3}{4} x + \frac{25}{4}$

Without calculus:

The tangent is perpendicular to the radius connecting $\left(0 , 0\right)$ and $\left(3 , 4\right)$.

The slope of the radius is $\frac{4}{3}$, so the slope of the tangent is $- \frac{3}{4}$