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