How do you write an equation of the line tangent to x2+y2=169 at the point...

Here we have the equation of a circle: ${\displaystyle x^2+y^2={13}^2}$
To determine the slope of a tangent to the circle at any point we need to use implicit differentiation.
${\displaystyle x^2+y^2={13}^2}$
${\displaystyle 2x+2y\cdot \frac{dy}{dx}=0}$
${\displaystyle \frac{dy}{dx}=-\frac{x}{y}}$
At the point (5,12), ${\displaystyle \frac{dy}{dx}=-\frac{5}{12}}$
So the tangent has a slope of ${\displaystyle -\frac{5}{12}}$ and passes through the point (5,12)
The equation of a line of slope m, passing through the point ${\displaystyle (x_1,y_1)}$ is:
${\displaystyle (y-y_1)=m(x-x_1)}$
The tangent would therefore have the equation:
${\displaystyle (y-12)=-\frac{5}{12}(x-5)}$
12y-144=-5x+25
12y=-5x+169
${\displaystyle y=\frac{1}{12}(-5x+169)}$

Explanation:
Differentiate implicitly:
${\displaystyle \frac{d}{dx}(x^2+y^2)=0}$
${\displaystyle 2x+2y\frac{dy}{dx}=0}$
${\displaystyle \frac{dy}{dx}=-\frac{x}{y}}$
So, for x=5, y=12:
${\displaystyle y^{\prime }\left(5\right)={\left[\frac{dy}{dx}\right]}_{5,12}=-\frac{5}{12}}$
The equation of the tangent line is:
${\displaystyle y=y_0+y^{\prime }\left(x_0\right)(x-x_0)=12-\frac{5}{12}(x-5)}$
${\displaystyle y=-\frac{5}{12}x+\frac{144+25}{12}}$
12y+5x=169