Question

A cellular phone network uses towers to

Answer 1

Answer:

Both $(6,\, 0)$ and $(8,\, 2)$ are inside this circular area.

$(6 - 5)^{2} + (0 - 1)^{2} = 2 < 22$.

$(8 - 5)^{2} + (2 - 1)^{2} = 10 < 22$.

Step-by-step explanation:

Equation for a circle in 2D, with center $(a,\, b)$ and radius $r$:

$(x - a)^{2} + (y - b)^{2} = r^{2}$.

Compare this expression with the one from this question:

$(x - 5)^{2} + (y - 1)^{2} = 22 = \left(\sqrt{22}\right)^{2}$.

Hence: $a = 5$, $b = 1$, and $r = \sqrt{22}$.

Therefore, $\! (5,\, 1)$ and $\sqrt{22}$ would be the center and the radius of the circle $(x - 5)^{2} + (y - 1)^{2} = 22$.

A point is inside a circle if and only the Euclidean distance between that point and the center of that circle is smaller than the radius of the circle. That is the same as requiring that the square of the Euclidean distance between these two points to be smaller than the square of the radius of the circle.

Formula for the Euclidean distance between $(x_1,\, y_1)$ and $(x_2,\, y_2)$:

$\displaystyle \sqrt{(x_1 - x_2)^{2} + (y_1 - y_2)^{2}}$.

The square of the Euclidean distance between these two points would be:

$\displaystyle (x_1 - x_2)^{2} + (y_1 - y_2)^{2}$.

Calculate the square of the distance between $(6,\, 0)$ and the center of the circle, $(5,\, 1)$.

$(6 - 5)^{2} + (0 - 1)^{2} = 2$.

The square of this distance is smaller than $22$, the square of the radius of this circle. Hence, the point $(6,\, 0)$ is inside this circle.

Similarly, calculate the square of the distance between $(8,\, 2)$ and the center of the circle, $(5,\, 1)$.

$(8 - 5)^{2} + (2 - 1)^{2} = 10$.

The square of this distance is smaller than $22$, the square of the radius of this circle. Hence, the point $(8,\, 2)$ is also inside this circle.

Notice that the point $(x,\, y)$ is on the 2D circle $(x - a)^{2} + (y - b)^{2} = r^{2}$ if and only if $x$ and $y$ satisfy the equation of that circle.

On the other hand, $(x,\, y)$ is inside this circle if and only $x$ and $y$ satisfy the inequality $(x - a)^{2} + (y - b)^{2} < r^{2}$.

Both $(6,\, 0)$ and $(8,\, 2)$ satisfy the inequality $(x - 5)^{2} + (y - 1)^{2} < 22$. Hence, both points are inside the circle $(x - 5)^{2} + (y - 1)^{2} = 22$.