Question 620·Medium·Math

Type $x^2+y^2-6x+8y+9=0$ as written, without adding $y=$ in front. Select the circle in Desmos. It marks the highest point $(3,0)$ and the lowest point $(3,-8)$; use those coordinates instead of estimating from the grid.

The top and bottom of a circle are the ends of a vertical diameter. A diameter passes through the center, so label the marked points $A$ and $B$ by typing $A=(3,0)$ and $B=(3,-8)$, then type $\operatorname{midpoint}(A,B)$. Click the plotted midpoint: Desmos shows $(3,-4)$. That is the center.

A radius reaches from the center to the circle, so type $\frac{\operatorname{distance}(A,B)}{2}$. Desmos displays $4$. That is half the diameter, giving a center of $(3,-4)$ and a radius of $4$. Choice D.

Move $+9$ to the right and group the $x$-terms and $y$-terms so each can become a square:

To complete the square, halve $-6$ to get $-3$, then square it to get $9$. Add $9$ to both sides so the equation stays balanced:

Halve $8$ to get $4$, then square it to get $16$. Add $16$ to both sides:

Each group is now a perfect square: $x^2-6x+9=(x-3)^2$ and $y^2+8y+16=(y+4)^2$. Rewrite the left side:

The signs inside the parentheses aren’t the center’s coordinates.

Type $-9+9+16$ in Desmos; it displays $16$. Replace the right side:

In standard form, $(x-h)^2+(y-k)^2=r^2$, the center is $(h,k)$. Here $x-3$ and $y+4$ become zero at $3$ and $-4$, so the center is $(3,-4)$. The right side is $r^2$, not $r$, so the radius is $\sqrt{16}=4$. Choice D.