Question 56·Medium·Geometry and Trigonometry

For each group, $x^2 - 6x$ and $y^2 + 4y$, what number can you add to make it a perfect-square trinomial?

Once you have an equation that looks like $(x - h)^2 + (y - k)^2 = r^2$, focus on the right-hand side to determine $r^2$ and then $r$.

In Desmos, type the equation x^2 + y^2 - 6x + 4y - 12 = 0 into an expression line. Desmos will graph the circle defined by this equation.

Click on the circle; Desmos will show a point at its center. Note the coordinates of this center point.

Click on a point where the circle intersects the grid (for example, where it meets a horizontal or vertical grid line). Use the distance from the center to that point; this distance is the radius. You can also use the Desmos distance formula tool distance((h,k),(x1,y1)) if you want a numerical confirmation.

Start with the given equation:

Group the $x$ terms together and the $y$ terms together, and move the constant to the other side:

For $x^2 - 6x$, take half of $-6$ (which is $-3$) and square it to get $9$. For $y^2 + 4y$, take half of $4$ (which is $2$) and square it to get $4$.

Add these values to both sides:

Now rewrite each trinomial as a square:

The standard form of a circle is

Comparing this with

you can see that $r^2 = 25$, so the radius is $r = 5$, which corresponds to choice B.