The equation of the tangent line to the curve x^2+y^2=169 at the point (5 -12) is

Point-slope form of a line:
y-y₀=m(x-x₀)

Data:
we have the point (5,-12); therefore:
x₀=5
y₀=-12

We have to calculate "m".
m=slope
The slope at the point x=5 will be.
m=f´(5)

We have to find the first derivative:
x²+y²=169
2x+2yy`=0
y´=-2x/2y

Therefore; the slope at the point x=5 is:
m=f´(5)=-2(5)/2(-12)=5/12

Point-slope form of this line:
y+12=5/12(x-5)
y+12=5/12 x-25/12
y=5/12 x-25/12  - 12
y=5/12 x -169/12  (slope-intercept form)

the equation of the tangent line to the curve at the point (5,-12) is:
y=5/12 x -169/12