Question

Consider the circle of radius 5 centered at (0,0), how do you find an equation of the line tangent to the circle at the point (3,4)?

Answer 1

Using calculus:

the equation of the circle is `x^2+y^2 = 25`

At the point `(3,4)`, we are on the upper semicircle, whose equation is: `y = sqrt(25-x^2)`

So `dy/dx = 1/(2sqrt(25-x^2)) (-2x) = (-x)/sqrt(25-x^2)`

At `(3,4)`, we get `dy/dx = (-3)/4`

So the slope of the line is `-3/4` and the line contains the point `(3,4)`.

`y=-3/4 x +25/4`

Without calculus:

The tangent is perpendicular to the radius connecting `(0,0)` and `(3,4)`.

The slope of the radius is `4/3`, so the slope of the tangent is `-3/4`