Question 172·Medium·Geometry and Trigonometry

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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 $(x-h)^2+(y-k)^2=r^2$ ?

When you see $x^2$ and $y^2$ with the same coefficient, aim to rewrite in $(x-h)^2+(y-k)^2=r^2$ . Group $x$ terms and $y$ terms, move the constant, then complete the square for each group. Be careful: $(x+4)^2$ means the center’s $x$ -coordinate is $-4$ , and the right side after balancing is $r^2$ .

Hints

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Recognize the circle pattern

If an equation has $x^2$ and $y^2$ (same coefficient) and no $xy$ term, it can be rewritten as a circle by completing the square.

Group and move the constant

Put the constant on the right side, then group the $x$ terms together and the $y$ terms together.

Complete the square twice

For $x^2+8x$ , add and subtract $(\tfrac{8}{2})^2$ . For $y^2-6y$ , add and subtract $(\tfrac{-6}{2})^2$ .

Desmos Guide

Graph the original equation

In Desmos, enter $x^2+y^2+8x-6y-11=0$ to graph the circle.

Graph an answer choice

Enter one answer choice, such as $(x+4)^2+(y-3)^2=36$ , and see whether it matches the original circle exactly.

Confirm by matching graphs

The correct choice is the one whose graph lies exactly on top of the graph from the original equation.

Step-by-step Explanation

Group $x$ -terms and $y$ -terms

Start with

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

Move the constant term to the other side and group:

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

Complete the square

Complete the square for each variable:

$x^2+8x=(x+4)^2-16$
$y^2-6y=(y-3)^2-9$

Substitute:

(x+4)^2-16+(y-3)^2-9=11.

Write the standard form

Add $16$ and $9$ to both sides:

(x+4)^2+(y-3)^2=11+16+9=36.

So the circle’s equation is $(x+4)^2+(y-3)^2=36$ .

Question 172·Medium·Geometry and Trigonometry

Strategy

Hints

Desmos Guide

Step-by-step Explanation

Study Hints

Strategy

Hints

Desmos Guide

Step-by-step Explanation