Question 772·Medium·Math

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A cell phone tower’s coverage boundary on a map is modeled by the equation

x^2+y^2+8x-6y-11=0.

Which choice gives the equation of this circle in the form $(x-h)^2+(y-k)^2=r^2$ ?

An expanded circle equation hides the center and radius that standard form shows. Graph the whole equation in Desmos and check its top point against the figure. Then use the marked center and a point on the boundary to write the equation. Use the opposite signs of the center coordinates inside the parentheses, and square the radius on the right.

Hints

Hint 1
You can graph the whole equation without solving for $y$ . Select the circle in Desmos and read its top point. Does its position agree with the center marked in the figure?
Hint 2
The marked point $P$ is level with the center. A radius goes from the center to the boundary, so their horizontal distance gives you the radius. What belongs on the right side of standard form: the radius or its square?
Hint 3
In standard form, $(h,k)$ is the center. The figure puts the center to the left of the $y$ -axis, so write $x-h$ carefully when $h$ is negative.

Step-by-step

Approach 1: Use the graph and the marked points

Step 1
Graph the given circle
Type the whole equation into Desmos, with no extra $y=$ . Select the circle; Desmos marks its top point at $(-4,9)$ . That point is directly above the center marked in the figure, so the graph agrees with what the figure shows.
Step 2
Find the center and radius
The figure labels $C=(-4,3)$ as the center and $P=(2,3)$ as a point on the circle. They're at the same height, so the distance between their $x$ -coordinates is the radius: $r=2-(-4)=6$ . The graphed top point is also $6$ units above $C$ .
Step 3
Write the circle in standard form
In $(x-h)^2+(y-k)^2=r^2$ , the center is $(h,k)$ . Substitute $h=-4$ , $k=3$ , and $r=6$ :
$(x-(-4))^2+(y-3)^2=6^2.$
Rewrite $x-(-4)$ as $x+4$ :
$(x+4)^2+(y-3)^2=6^2.$
Square the radius:
$(x+4)^2+(y-3)^2=36.$
That is the tower's coverage boundary in the requested form. Choice A.

Approach 2: Complete both squares

Step 1
Move the constant
You can recover standard form from the equation without the figure. Move $-11$ to the right by adding $11$ to both sides:
$x^2+y^2+8x-6y=11.$
Step 2
Group the terms by variable
Put the $x$ -terms together and the $y$ -terms together so each group can become one squared parenthesis:
$(x^2+8x)+(y^2-6y)=11.$
Step 3
Complete and balance both squares
To complete a square, halve the coefficient of the unsquared variable, then square that half. Half of $8$ is $4$ , so add $4^2=16$ to the $x$ -group. Half of $-6$ is $-3$ , so add $(-3)^2=9$ to the $y$ -group. Add both numbers to the right too:
$(x^2+8x+16)+(y^2-6y+9)=11+16+9.$
Type $11+16+9$ in Desmos; it shows $36$ for the new right side.
Step 4
Rewrite the squares
The completed groups are $(x+4)^2$ and $(y-3)^2$ , so the same coverage boundary is
$(x+4)^2+(y-3)^2=36.$
Choice A.

Question 772·Medium·Math

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