# How do you write an equation for a circle tangent to the line 5x + y = 3 at the point (2,-7) and the center is on the line x – 2y= 19?

Oct 25, 2016

Please see the explanation. The equation is:
${\left(x - 7\right)}^{2} + {\left(y - - 6\right)}^{2} = {\left(\sqrt{26}\right)}^{2}$

#### Explanation:

Write the equation of the tangent line in slope-intercept form so that we may observe its slope.

$y = - 5 x + 3$

The slope $m = - 5$

Because radius of the circle at point (2, -7) will be perpendicular to the tangent line, the slope of the radius is the negative reciprocal of the slope of the tangent line, $\frac{1}{5}$, Write the equation of the line through the point with this slope:

$- 7 = \frac{1}{5} \left(2\right) + b$

$b = - 7 - \frac{2}{5}$

$b = - \frac{37}{5}$

$y = \frac{1}{5} x - \frac{37}{5}$ [1]

The center $\left(h , k\right)$ will be at the intersection of equation [1] and the line:

$x - 2 y = 19$ [2]

Substitute equation [1] into equation [2]:

$x - 2 \left(\frac{1}{5} x - \frac{37}{5}\right) = 19$

$x - \frac{2}{5} x + \frac{74}{5} = 19$

$\frac{3}{5} x = \frac{21}{5}$

$x = 7$

Substituting for x in equation [2]:

$7 - 2 y = 19$

$- 2 y = 12$

$y = - 6$

Substitute the center $\left(7 , - 6\right)$ and the point $\left(2 , - 7\right)$ into the equation of a circle:

${\left(2 - 7\right)}^{2} + {\left(- 7 - - 6\right)}^{2} = {r}^{2}$

#r = sqrt(26)