Question

A circle is centered at J(3, 3) and has a radius of 12.

Answer 1

Answer:

$(-6,\, -5)$ is outside the circle of radius of $12$ centered at $(3,\, 3)$.

Step-by-step explanation:

Let $J$ and $r$ denote the center and the radius of this circle, respectively. Let $F$ be a point in the plane.

Let $d(J,\, F)$ denote the Euclidean distance between point $J$ and point $F$.

In other words, if $J$ is at $(x_j,\, y_j)$ while $F$ is at $(x_f,\, y_f)$, then $\displaystyle d(J,\, F) = \sqrt{(x_j - x_f)^{2} + (y_j - y_f)^{2}}$.

Point $F$ would be inside this circle if $d(J,\, F) < r$. (In other words, the distance between $F\!$ and the center of this circle is smaller than the radius of this circle.)

Point $F$ would be on this circle if $d(J,\, F) = r$. (In other words, the distance between $F\!$ and the center of this circle is exactly equal to the radius of this circle.)

Point $F$ would be outside this circle if $d(J,\, F) > r$. (In other words, the distance between $F\!$ and the center of this circle exceeds the radius of this circle.)

Calculate the actual distance between $J$ and $F$:

$\begin{aligned}d(J,\, F) &= \sqrt{(x_j - x_f)^{2} + (y_j - y_f)^{2}}\\ &= \sqrt{(3 - (-6))^{2} + (3 - (-5))^{2}} \\ &= \sqrt{145} \end{aligned}$.

On the other hand, notice that the radius of this circle, $r = 12 = \sqrt{144}$, is smaller than $d(J,\, F)$. Therefore, point $F$ would be outside this circle.