# How do you write the equation of the line that is tangent to the circle (x + 6)^2 + (y + 4)^2 = 25 at the point (-9, -8)?

Jul 12, 2016

We will have to differentiate the function through implicit differentiation to find this one.

Let's first distribute:

${x}^{2} + 12 x + 36 + {y}^{2} + 8 y + 16 = 25$

${x}^{2} + {y}^{2} + 12 x + 8 y + 52 - 25 = 0$

${x}^{2} + {y}^{2} + 12 x + 8 y + 27 = 0$

${x}^{2} + {y}^{2} + 12 x + 8 y = - 27$

We start by differentiating both sides:

$\frac{d}{\mathrm{dx}} \left({x}^{2} + {y}^{2} + 12 x + 8 y\right) = \frac{d}{\mathrm{dx}} \left(- 27\right)$

$2 x + 2 y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + 12 + 8 \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

$2 y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + 8 \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = - 12 - 2 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(2 y + 8\right) = - 12 - 2 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 12 - 2 x}{2 y + 8}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 \left(6 + x\right)}{2 \left(y + 4\right)}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x + 6}{y + 4}$

The slope of the tangent is given by inserting our given point, $\left(x , y\right)$, into the derivative.

Hence, ${m}_{\text{tangent}} = - \frac{- 9 + 6}{- 8 + 4} = - \left(- \frac{3}{-} 4\right) = - \frac{3}{4}$

The equation of the tangent can now be found by using point-slope form, since this will be a straight line.

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - \left(- 8\right) = - \frac{3}{4} \left(x - \left(- 9\right)\right)$

$y + 8 = - \frac{3}{4} x - \frac{27}{4}$

$y = - \frac{3}{4} x - \frac{27}{4} - 8$

$y = - \frac{3}{4} x - \frac{59}{4}$

So, the equation of the tangent to $y = {\left(x + 6\right)}^{2} + {\left(y + 4\right)}^{2} = 25$ at the point $\left(- 9 , - 8\right)$ is $y = - \frac{3}{4} x - \frac{59}{4}$.

Hopefully this helps!