Animated Solution for Mathematics - Circles: If one of the diameters of the circle $x^2+y^2-2\sqrt{2}x-6\sqrt{2}y+14=0$ is a chord of the circle $(x-2\sqrt{2}{)}^2+(y-2\sqrt{2}{)}^2=r^2$, then the value of $r^2$ is equal to ____.

We begin with the first circle, $C_1$, defined by the equation $x^2+y^2-2\sqrt{2}x-6\sqrt{2}y+14=0$. By comparing this to the general form $x^2+y^2+2gx+2fy+c=0$, we identify the coefficients: $2g=-2\sqrt{2}\Rightarrow g=-\sqrt{2}$ $2f=-6\sqrt{2}\Rightarrow f=-3\sqrt{2}$

The center $O_1$ is given by $(-g,-f)$, which yields $O_1=(\sqrt{2},3\sqrt{2})$.

To find the radius $r_1$, we use the standard formula:

Substituting our values:

Next, we examine $C_2$, given by the equation $(x-2\sqrt{2}{)}^2+(y-2\sqrt{2}{)}^2=r^2$. This is already in the standard form $(x-h{)}^2+(y-k{)}^2=r^2$.

We can immediately identify the center $O_2=(2\sqrt{2},2\sqrt{2})$. The radius of this circle is $r$, which remains our unknown variable to solve.

The problem states that a diameter of $C_1$ is a chord of $C_2$. Since the diameter of $C_1$ passes through its center $O_1$, the chord of $C_2$ must also pass through $O_1$.

In any circle, the perpendicular distance from the center to a chord bisects that chord. Here, the line segment connecting the centers $O_1$ and $O_2$ is perpendicular to the chord. This forms a right-angled triangle $△O_2{O}_{1}P$, where $P$ is an endpoint of the chord on the circumference of $C_2$.

In the right-angled triangle $△O_2{O}_{1}P$, the hypotenuse is the radius of $C_2$ (which is $r$), one leg is the radius of $C_1$ ($r_1=\sqrt{6}$), and the other leg is the distance $d$ between the centers $O_1$ and $O_2$.

First, we calculate the square of the distance $d$ between $O_1(\sqrt{2},3\sqrt{2})$ and $O_2(2\sqrt{2},2\sqrt{2})$:

Applying the Pythagorean theorem, $r^2=r_1^2+d^2$:

The final value is $r^2=10$. By visualizing the geometric relationship between the centers and the chord, we have arrived at the solution with precision and elegance.