Sarah-Louise Prince

2022-02-15

How do you write an equation of the line tangent to ${x}^{2}+{y}^{2}=169$ at the point (5,12)?

Kathleen Mcpherson

Beginner2022-02-16Added 9 answers

Here we have the equation of a circle: $x}^{2}+{y}^{2}={13}^{2$

To determine the slope of a tangent to the circle at any point we need to use implicit differentiation.

$x}^{2}+{y}^{2}={13}^{2$

$2x+2y\cdot \frac{dy}{dx}=0$

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

At the point (5,12),$\frac{dy}{dx}=-\frac{5}{12}$

So the tangent has a slope of$-\frac{5}{12}$ and passes through the point (5,12)

The equation of a line of slope m, passing through the point$({x}_{1},{y}_{1})$ is:

$(y-{y}_{1})=m(x-{x}_{1})$

The tangent would therefore have the equation:

$(y-12)=-\frac{5}{12}(x-5)$

12y-144=-5x+25

12y=-5x+169

$y=\frac{1}{12}(-5x+169)$

To determine the slope of a tangent to the circle at any point we need to use implicit differentiation.

At the point (5,12),

So the tangent has a slope of

The equation of a line of slope m, passing through the point

The tangent would therefore have the equation:

12y-144=-5x+25

12y=-5x+169

surgescasjag

Beginner2022-02-17Added 10 answers

Explanation:

Differentiate implicitly:

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

$2x+2y\frac{dy}{dx}=0$

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

So, for x=5, y=12:

$y}^{\prime}\left(5\right)={\left[\frac{dy}{dx}\right]}_{5,12}=-\frac{5}{12$

The equation of the tangent line is:

$y={y}_{0}+{y}^{\prime}\left({x}_{0}\right)(x-{x}_{0})=12-\frac{5}{12}(x-5)$

$y=-\frac{5}{12}x+\frac{144+25}{12}$

12y+5x=169

Differentiate implicitly:

So, for x=5, y=12:

The equation of the tangent line is:

12y+5x=169