Equation of Tangents to Circle from Point

Theorem

Let $\CC$ be a circle embedded in the Cartesian plane of radius $r$ with its center located at the origin.

Let $P = \tuple {x_0, y_0}$ be a point in the plane of $\CC$ which is outside $\CC$.

The tangents to $\CC$ which pass through $P$ can be described using the equation:

$\paren {x x_0 + y y_0 - r^2}^2 = \paren {x^2 + y^2 - r^2} \paren { {x_0}^2 + {y_0}^2 - r^2}$

Proof

From Equation of Circle center Origin, $\CC$ can be described as:

$x^2 + y^2 = r^2$

Let $\LL$ be an arbitrary straight line through $P$ which intersects $\CC$ at $U$ and $V$.

Let $Q = \tuple{x, y}$ be an arbitrary point on $\LL$.

Let $k$ be the position-ratio of one of the points $U$ and $V$ with respect to $P$ and $Q$.

By Joachimsthal's equation:

$(1): \quad k^2 \paren {x^2 + y^2 - r^2} + 2 k \paren {x x_0 + y y_0 - r^2} + \paren { {x_0}^2 + {y_0}^2 - r^2} = 0$

which is a quadratic in $k$.

When $\LL$ is tangent to $\CC$, the points $U$ and $V$ coincide.

Hence $(1)$ has equal roots.

From Solution to Quadratic Equation, it follows that the discriminant of $(1)$ is zero.

That is:

$\paren {x x_0 + y y_0 - r^2}^2 - \paren {x^2 + y^2 - r^2} \paren { {x_0}^2 + {y_0}^2 - r^2} = 0$

$\blacksquare$

Sources

• 1933: D.M.Y. Sommerville: Analytical Conics (3rd ed.) ... (previous) ... (next): Chapter $\text {III}$. The Circle: $12$. Equation of the tangents to a given circle from a given point