Theorem on the lengths of a tangent and a secant; the product of the entire secant and its external part. Trigonometric ratios in a right-angled triangle

A circle and a line are tangent at point $M$. Perpendiculars from points $A$ and $B$ of this circle to the line are equal to $a$ and $b$ respectively. Find the distance from point $M$ to the line $A B$.

#
Apply the theorem of the tangent and secant.

## Solution

If line $A B$ is parallel to the tangent, then everything is obvious.

Let the specified tangent and line $A B$ intersect at point $K$ at an angle $\alpha$, and let $x$ be the desired segment. Then

$$
\sin \alpha=\frac{b}{B K}=\frac{x}{M K}=\frac{a}{A K}
$$

Multiplying the equalities term by term

$$
\frac{a}{A K}=\frac{x}{M K}, \frac{b}{B K}=\frac{x}{M K}
$$

we get:

$$
\frac{a b}{A K \cdot B K}=\frac{x^{2}}{M K^{2}}
$$

Since $A K \cdot B K=M K^{2}$, then $x^{2}=a b$. Therefore, $x=\sqrt{a b}$.

If line $A B$ is parallel to the tangent, then everything is obvious.

Let the specified tangent and line $A B$ intersect at point $K$ at an angle $\alpha$, and let $x$ be the desired segment. Then

$$
\sin \alpha=\frac{b}{B K}=\frac{x}{M K}=\frac{a}{A K}
$$

Multiplying the equalities term by term

$$
\frac{a}{A K}=\frac{x}{M K}, \frac{b}{B K}=\frac{x}{M K}
$$

we get:

$$
\frac{a b}{A K \cdot B K}=\frac{x^{2}}{M K^{2}}
$$

Since $A K \cdot B K=M K^{2}$, then $x^{2}=a b$. Therefore, $x=\sqrt{a b}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_96a7587f2ba0a7b3aee2g-39.jpg?height=380&width=809&top_left_y=1708&top_left_x=630)

Answer

$\sqrt{a b}$.