Answer:
[tex](-6,\, -5)[/tex] is outside the circle of radius of [tex]12[/tex] centered at [tex](3,\, 3)[/tex].
Step-by-step explanation:
Let [tex]J[/tex] and [tex]r[/tex] denote the center and the radius of this circle, respectively. Let [tex]F[/tex] be a point in the plane.
Let [tex]d(J,\, F)[/tex] denote the Euclidean distance between point [tex]J[/tex] and point [tex]F[/tex].
In other words, if [tex]J[/tex] is at [tex](x_j,\, y_j)[/tex] while [tex]F[/tex] is at [tex](x_f,\, y_f)[/tex], then [tex]\displaystyle d(J,\, F) = \sqrt{(x_j - x_f)^{2} + (y_j - y_f)^{2}}[/tex].
Point [tex]F[/tex] would be inside this circle if [tex]d(J,\, F) < r[/tex]. (In other words, the distance between [tex]F\![/tex] and the center of this circle is smaller than the radius of this circle.)
Point [tex]F[/tex] would be on this circle if [tex]d(J,\, F) = r[/tex]. (In other words, the distance between [tex]F\![/tex] and the center of this circle is exactly equal to the radius of this circle.)
Point [tex]F[/tex] would be outside this circle if [tex]d(J,\, F) > r[/tex]. (In other words, the distance between [tex]F\![/tex] and the center of this circle exceeds the radius of this circle.)
Calculate the actual distance between [tex]J[/tex] and [tex]F[/tex]:
[tex]\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}[/tex].
On the other hand, notice that the radius of this circle, [tex]r = 12 = \sqrt{144}[/tex], is smaller than [tex]d(J,\, F)[/tex]. Therefore, point [tex]F[/tex] would be outside this circle.
