Question 656·Medium·Math

0 / 2000

What is the radius of the circle defined by the equation

x^{2}+y^{2}-6x+4y-12 = 0

An expanded circle equation has squared $x$ and $y$ terms alongside separate $x$ and $y$ terms. For a numerical radius, graph the whole equation in Desmos and click the circle’s highest and lowest points. Their vertical separation is the diameter, or full distance across the circle. Halve that distance; reporting the full span gives the diameter instead.

Hints

Hint 1
You can graph an expanded circle equation without rearranging it. Enter the entire equality in Desmos and select the circle to reveal its highest and lowest points. What distance do those points span?
Hint 2
A diameter is the full distance across a circle through its center. The top and bottom points line up vertically, so subtract their $y$ -coordinates to find the vertical distance between them.
Hint 3
A radius runs from the center to the circle, so it is half a diameter. Once you have the distance from the top point to the bottom point, how much of it do you need?

Step-by-step

Approach 1: Read the diameter from the graph
Step 1
Find the circle’s top and bottom points
Type the complete equation $x^{2}+y^{2}-6x+4y-12=0$ in Desmos, without adding $y=$ in front. Click the circle. Desmos marks its highest and lowest points at $(3,3)$ and $(3,-7)$ .
Step 2
Find the full vertical span
The points share $x=3$ , so the distance between their $y$ -coordinates is a diameter, the full distance across the circle through its center. Subtract the lower coordinate from the upper one:
$d=3-(-7)=10.$
Step 3
Halve the diameter
A radius goes from the center to the circle. The radius is half the full diameter. Divide by $2$ :
$r=\frac{10}{2}=5.$
The circle’s radius is $5$ . Choice B.

Approach 2: Complete the squares
Step 1
Move the constant
To uncover the circle’s standard form, start by adding $12$ to both sides:
$x^2+y^2-6x+4y=12.$
Step 2
Complete the $x$ square
Half of the coefficient $-6$ is $-3$ , and its square is $9$ . Add $9$ to both sides so the $x$ terms can form a square:
$(x^2-6x+9)+y^2+4y=12+9.$
Add the right side:
$(x^2-6x+9)+y^2+4y=21.$
Step 3
Complete the $y$ square
Half of the coefficient $4$ is $2$ , and its square is $4$ . Add $4$ to both sides so the $y$ terms can form a square:
$(x^2-6x+9)+(y^2+4y+4)=21+4.$
Add the right side:
$(x^2-6x+9)+(y^2+4y+4)=25.$
Step 4
Rewrite the circle equation
Each three-term group is now a binomial square: $x^2-6x+9=(x-3)^2$ and $y^2+4y+4=(y+2)^2$ . Rewrite the equation:
$(x-3)^2+(y+2)^2=25.$
Step 5
Read the radius
In a circle’s standard form, $(x-h)^2+(y-k)^2=r^2$ , the right side is the squared radius, not the radius. Take its positive square root:
$r=\sqrt{25}=5.$
The circle’s radius is $5$ . Choice B.

Question 656·Medium·Math

Strategy

Hints

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Step-by-step

Lessons that teach this