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

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 $\left({x}_{1},{y}_{1}\right)$ is:
$\left(y-{y}_{1}\right)=m\left(x-{x}_{1}\right)$
The tangent would therefore have the equation:
$\left(y-12\right)=-\frac{5}{12}\left(x-5\right)$
12y-144=-5x+25
12y=-5x+169
$y=\frac{1}{12}\left(-5x+169\right)$

surgescasjag

Explanation:
Differentiate implicitly:
$\frac{d}{dx}\left({x}^{2}+{y}^{2}\right)=0$
$2x+2y\frac{dy}{dx}=0$
$\frac{dy}{dx}=-\frac{x}{y}$
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:
$y={y}_{0}+{y}^{\prime }\left({x}_{0}\right)\left(x-{x}_{0}\right)=12-\frac{5}{12}\left(x-5\right)$
$y=-\frac{5}{12}x+\frac{144+25}{12}$
12y+5x=169

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