Equation of Tangent to Circle Centered at Origin/Proof 1
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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 Equation of Straight Line Tangent to Circle we have that for a general circle of radius $r$ and center $\tuple {a, b}$:
- $y - y_1 = \dfrac {a - x_1} {y_1 - b} \paren {x - x_1}$
is the equation of a tangent $\TT$ to $\CC$ passing through $\tuple {x_1, y_1}$.
Setting the center to $\tuple {0, 0}$:
\(\ds y - y_1\) | \(=\) | \(\ds -\dfrac {x_1} {y_1} \paren {x - x_1}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds y_1 \paren {y - y_1}\) | \(=\) | \(\ds -x_1 \paren {x - x_1}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x x_1 + y y_1\) | \(=\) | \(\ds x_1^2 + y_1^2\) | |||||||||||
\(\ds \) | \(=\) | \(\ds r^2\) | as $\tuple {x_1, y_1}$ is on $\CC$ |
$\blacksquare$