[Theorem of the tangent and secant; the product of the entire secant and its external part [ Inscribed angle, based on the diameter

A circle with a diameter equal to $\sqrt{10}$ passes through the adjacent vertices $A$ and $B$ of the rectangle $A B C D$.

The length of the tangent drawn from point $C$ to the circle is $3, A B=1$. Find all possible values that the length of side $B C$ can take.

#
Consider two possible cases, and in each of them apply the theorem of the tangent and the secant.

## Solution

Note that vertex $C$ is located outside the circle. Let $CK$ be the given tangent ($K$ - the point of tangency). If the circle has no common points with the given rectangle, except for points $A$ and $B$, then, extending the segment $CB$ to intersect the circle at point $P$, we obtain a right triangle $ABP$, with the hypotenuse $AP$ being the diameter of the circle. Therefore,

$$
BP = \sqrt{AP^2 - AB^2} = \sqrt{10 - 1} = 3
$$

By the theorem of the tangent and the secant,

$$
BC(BC + BP) = CK^2, \text{ or } BC(BC + 3) = 9.
$$

From this, we find that

$$
BC = \frac{3}{2}(\sqrt{5} - 1).
$$

If the circle intersects the rectangle at points other than $A$ and $B$, then the corresponding equation is

$$
BC(BC - 3) = 9.
$$

Its root is

$$
BC = \frac{3}{2}(\sqrt{5} + 1).
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-24.jpg?height=466&width=1238&top_left_y=1884&top_left_x=408)

## Answer

$\frac{3}{2}(\sqrt{5} \pm 1)$.