Question

Consider the circle of radius 5 centered at (0, 0). Find an equation of the line tangent to the circle at the point (3, 4) in slope intercept form.

Answer 1

`\displaystyle y= -(3)/(4) x + (25)/(4)`.

Explanation:

The equation of a circle of radius
`5` centered at
`(0,0)` is:

`x^(2) + y^(2) = 5^(2)`.

`x^(2) + y^(2) = 25`.

Differentiate implicitly with respect to
`x` to find the slope of tangents to this circle.

`\displaystyle (d)/(dx)[x^(2) + y^(2)] = (d)/(dx)[25]`

`\displaystyle (d)/(dx)(x^(2)) + (d)/(dx)(y^(2)) = 0`.

Apply the power rule and the chain rule. Treat
`y` as a function of
`x`,
`f(x)`.

`\displaystyle (d)/(dx)(x^(2)) + (d)/(dx)(f(x))^(2) = 0`.

`\displaystyle (d)/(dx)(2x) + (d)/(dx)(2f(x)\cdot f^(\prime)(x)) = 0`.

That is:

`\displaystyle (d)/(dx)(2x) + (d)/(dx)\left(2y \cdot (dy)/(dx)\right) = 0`.

Solve this equation for
`\displaystyle (dy)/(dx)`:

`\displaystyle (dy)/(dx) = -(x)/(y)`.

The slope of the tangent to this circle at point
`(3, 4)` will thus equal

`\displaystyle (dy)/(dx) = -(3)/(4)`.

Apply the slope-point of a line in a cartesian plane:

`y - y_0 = m(x - x_0)`, where

• 
  `m` is the gradient of this line, and
• 
  `(x_0, y_0)` are the coordinates of a point on that line.

For the tangent line in this question:

• 
  `\displaystyle m = -(3)/(4)`,
• 
  `(x_0, y_0) = (3, 4)`.

The equation of this tangent line will thus be:

`\displaystyle y - 4 = -(3)/(4) (x - 3)`.

That simplifies to

`\displaystyle y= -(3)/(4) x + (25)/(4)`.