Find the equation of the line that is tangent to the circle x2 + y2 = 36 at point (4, -5).

Although we can only use the x-coordinate to find the slope of the tangent that actually touches the circle but that point wont be (4,-5) it'd be some other point (4,a)

But in order to still solve the for the tangent that satisfies (4,-5) and the circle. here's the solution:

To find the equation of the tangent, we need to find the slope of the line at the point (4,-5).

To find the slope of the circle, we need to first differentiate it!

given equation is:

[tex]x^2+y^2=36[/tex]

now you can either make y the subject of this equation differentiate then OR you can differentiate it directly. I'm gonna go with the latter since its far easier also we can use both coordinates (4,-5) to find the slope instead of only the x-coordinate.

[tex]\dfrac{d}{x}(x^2+y^2)=\dfrac{d}{dx}(36)[/tex]

[tex]2x+2y\dfrac{dy}{dx})=0[/tex]

[tex]\dfrac{dy}{dx})=\dfrac{-2x}{2y}[/tex]

[tex]\dfrac{dy}{dx})=-\dfrac{x}{y}[/tex]

this is the equation of the slope of our circle at any point (x,y). to find the slop at (4,-5). we'll plug in these values.

[tex]\dfrac{dy}{dx})=-\dfrac{4}{-5}[/tex]

[tex]\dfrac{dy}{dx})=\dfrac{4}{5}[/tex]

this is also the slope of the tangent at (4,-5), hence,[tex]m=\dfrac{4}{5}[/tex]

now to find the equation of the tangent: (we'll use the general equation of the line formula)

[tex](y-y_1) = m(x-x_1)[/tex]

here, m = 4/5 and (x1,y1) =(4,-5)

[tex](y-(-5)) = \dfrac{4}{5}(x-4)[/tex]

[tex]y = \dfrac{4}{5}x-\dfrac{4(4)}{5}-5[/tex]

[tex]y = \dfrac{4}{5}x-\dfrac{41}{5}[/tex]

This is the equation of the tangent to the circle!