Animated Solution for Mathematics - Circles: Let the tangents drawn from the origin to the circle, $x^2+y^2-8x-4y+16=0$ touch it at the points $A$ and $B$. The $(AB{)}^2$ is equal to :

We are given the circle equation $x^2+y^2-8x-4y+16=0$. To understand its geometry, we first complete the square for the $x$ and $y$ terms.

Grouping the terms, we have $(x^2-8x)+(y^2-4y)=-16$. Adding the necessary constants to both sides, we obtain:

This simplifies to the standard form:

From this, we identify the center $C$ at $(4,2)$ and the radius $r=2$.

We consider the tangents drawn from the origin $O(0,0)$ to the circle. The length $L$ of the tangent from an external point to a circle is given by $L=\sqrt{S_1}$, where $S_1$ is the value of the circle's equation evaluated at that point.

Substituting $(0,0)$ into the original equation:

Thus, the length of the tangent segment from the origin to the point of contact is exactly $4$ units.

Connecting the points of contact $A$ and $B$ forms the chord of contact. We utilize the geometric relationship between the radius $r$, the tangent length $L$, and the distance $d$ from the center to the origin, where $d=\sqrt{4^2+2^2}=\sqrt{20}$.

The length of the chord of contact $AB$ is given by the formula:

This formula arises from the properties of the right-angled triangle formed by the center, the point of contact, and the external point.

We substitute our known values $L=4$ and $r=2$ into the chord length formula:

The problem asks for the square of the length of the chord, $(AB{)}^2$. Calculating this:

The final result is $12.8$ or $\frac{64}{5}$.