Equation of Tangent to Circle Centered at Origin/Proof 2

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Theorem

Let $\CC$ be a circle whose radius is $r$ and whose center is at the origin of a Cartesian plane.

Let $P = \tuple {x_1, y_1}$ be a point on $\CC$.

Let $\TT$ be a tangent to $\CC$ passing through $P$.


Then $\TT$ can be defined by the equation:

$x x_1 + y y_1 = r^2$


Proof

From the slope-intercept form of a line, the equation of a line passing through $P$ is:

$y - y_1 = \mu \paren {x - x_1}$

If this line passes through another point $\tuple {x_2, y_2}$ on $\CC$, the slope of the line is given by:

$\mu = \dfrac {y_2 - y_1} {x_2 - x_1}$


Because $P$ and $Q$ both lie on $\CC$, we have:

\(\ds x_1^2 + y_1^2\) \(=\) \(\ds r^2 = x_2^2 + y_2^2\)
\(\ds \leadsto \ \ \) \(\ds y_2^2 - y_1^2\) \(=\) \(\ds x_2^2 - x_1^2\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {y_2 - y_1} {x_2 - x_1}\) \(=\) \(\ds -\dfrac {x_1 + x_2} {y_1 + y_2}\)

As $Q$ approaches $P$, we have that $y_2 \to y_1$ and $x_2 \to x_1$.

The limit of the slope is therefore:

$-\dfrac {2 x_1} {2 y_1} = -\dfrac {x_1} {y_1}$


The equation of the tangent $\TT$ to $\CC$ passing through $\tuple {x_1, y_1}$ is therefore:

\(\ds y - y_1\) \(=\) \(\ds -\dfrac {x_1} {y_1} \paren {x - x_1}\)
\(\ds \leadsto \ \ \) \(\ds y_1 \paren {y - y_1}\) \(=\) \(\ds -x_1 \paren {x - x_1}\)
\(\ds \leadsto \ \ \) \(\ds x x_1 + y y_1\) \(=\) \(\ds x_1^2 + y_1^2\)
\(\ds \) \(=\) \(\ds r^2\) as $\tuple {x_1, y_1}$ is on $\CC$

$\blacksquare$


Sources

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