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
- 1933: D.M.Y. Sommerville: Analytical Conics (3rd ed.) ... (previous) ... (next): Chapter $\text {III}$. The Circle: $2$. The tangent at a given point