write the equation of the line that is tangent at the given point (x-2)^2+(y-4)^2=289;(-15,4)

Question

Using slope-point formula

Answer 1

derivative(slope):   2(x-2)+2(y-4)(y')=0

y'=-(x-2)/(y-4)

so at (-15,4) slope(y') = -(-15-2)/(4-4)=17/0

therefore the line is vertical (infinite rise over no run)

so the tangent line is a vertical line at the given y coordinate,

so the line would be x=-15