For a point $(x,y)$, think about what the value of $(x-3)^2+(y+2)^2$ tells you in relation to the circle’s radius squared, which is $49$.

Substitute the $x$- and $y$-coordinates of each answer choice into $(x-3)^2+(y+2)^2$ and compare each result to $49$. Which one gives a value greater than $49$?

In Desmos, enter the equation (x-3)^2 + (y+2)^2 = 49 to graph the circle with center $(3,-2)$ and radius $7$.

On separate lines, enter each point as an ordered pair: (4,-2), (10,-2), (-4,5), and (3,5). Desmos will display each as a point on the graph.

Look at where each point is in relation to the circle: identify which point is clearly beyond the circle’s boundary, not on or inside it. That point is the one that lies outside the circle.

You can also type the expression (x-3)^2 + (y+2)^2 and use a table with each $x$- and $y$-value to see the numerical result for each point; the point whose value is greater than 49 is the one outside the circle.

The given circle is

A circle in the form $(x-h)^2+(y-k)^2=r^2$ has center $(h,k)$ and radius $r$.

So here:

• $h=3$, $k=-2$, so the center is $(3,-2)$.
• $r^2=49$, so the radius is $r=7$.

For any point $(x,y)$, the left side of the equation

represents the square of the distance from $(x,y)$ to the center $(3,-2)$.

• If $(x-3)^2+(y+2)^2<49$, the point is inside the circle.
• If $(x-3)^2+(y+2)^2=49$, the point is on the circle.
• If $(x-3)^2+(y+2)^2>49$, the point is outside the circle.

We want the point whose value is greater than 49.

Compute $(x-3)^2+(y+2)^2$ for each point:

• A) $(4,-2)$:

which is less than $49$ (inside).

• B) $(10,-2)$:

which equals $49$ (on the circle).

• C) $(-4,5)$:

which is greater than $49$ (candidate for outside).

• D) $(3,5)$:

which equals $49$ (on the circle).

Only the point whose value of $(x-3)^2+(y+2)^2$ is greater than 49 lies outside the circle. From the calculations, $(-4,5)$ gives $98>49$, so the point outside the circle is $(-4,5)$.