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

Answer:

$\displaystyle y= -\frac{3}{4} x + \frac{25}{4}$.

Step-by-step 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.

$\displaystyle \frac{d}{dx}[x^{2} + y^{2}] = \frac{d}{dx}[25]$

$\displaystyle \frac{d}{dx}(x^{2}) + \frac{d}{dx}(y^{2}) = 0$.

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

$\displaystyle \frac{d}{dx}(x^{2}) + \frac{d}{dx}(f(x))^{2} = 0$.

$\displaystyle \frac{d}{dx}(2x) + \frac{d}{dx}(2f(x)\cdot f^{\prime}(x)) = 0$.

That is:

$\displaystyle \frac{d}{dx}(2x) + \frac{d}{dx}\left(2y \cdot \frac{dy}{dx}\right) = 0$.

Solve this equation for $\displaystyle \frac{dy}{dx}$:

$\displaystyle \frac{dy}{dx} = -\frac{x}{y}$.

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

$\displaystyle \frac{dy}{dx} = -\frac{3}{4}$.

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:

• $\displaystyle m = -\frac{3}{4}$,
• $(x_0, y_0) = (3, 4)$.

The equation of this tangent line will thus be:

$\displaystyle y - 4 = -\frac{3}{4} (x - 3)$.

That simplifies to

$\displaystyle y= -\frac{3}{4} x + \frac{25}{4}$.