Question 788·Medium·Math

0 / 2000

A marine research buoy sends a signal to any boat within its circular range. On a coordinate map, the boundary of the signal range is modeled by

x^2+y^2-10x+4y=20.

Which choice is the equation of this circle in standard form?

An expanded circle equation has $x^2$ and $y^2$ , but its center and radius are hidden. Graph the whole equation in Desmos and use the top and bottom points to find both. Then write standard form, $(x-h)^2+(y-k)^2=r^2$ . The right side is the radius squared, not necessarily the constant in the expanded equation.

Hints

Hint 1
Type the whole equation into Desmos without adding $y=$ . A circle can have two $y$ -values for one $x$ -value, so graph the relation as written. Where are its top and bottom points?
Hint 2
The top and bottom points are the ends of a vertical diameter, a line through the circle's center. The center is halfway between those points. What are its coordinates?
Hint 3
A radius runs from the center to the circle's edge. Find the distance from the center to the top point, then put its square on the right side of $(x-h)^2+(y-k)^2=r^2$ .

Step-by-step

Approach 1: Graph the boundary, then write standard form

Step 1
Find the circle's top and bottom
Type $x^2+y^2-10x+4y=20$ as written in Desmos. Click the circle to see its highest point, $(5,5)$ , and lowest point, $(5,-9)$ . These are the ends of the vertical diameter drawn in the figure.
Step 2
Locate the center
The center is halfway between the diameter's endpoints. Both have $x=5$ . Type $\frac{5+(-9)}{2}$ in Desmos; it prints $-2$ . So the center is $(5,-2)$ .
Step 3
Find the radius
The radius reaches from the center $(5,-2)$ to the top point $(5,5)$ , so its length is $5-(-2)=7$ . The full diameter runs from top to bottom; don't use that longer distance for $r$ .
Step 4
Put the center and radius into standard form
In standard form, $(x-h)^2+(y-k)^2=r^2$ , the center is $(h,k)$ . Substitute the center and radius you found:
$(x-5)^2+(y-(-2))^2=7^2.$
Step 5
Simplify the equation
Simplify the negative $y$ -coordinate and square the radius:
$(x-5)^2+(y+2)^2=49.$
That's the buoy's signal boundary in standard form. Choice A.

Approach 2: Complete both squares from the equation

Step 1
Complete the x-square
Half of the $x$ -term's coefficient, $-10$ , is $-5$ ; its square is $25$ . Adding $25$ turns $x^2-10x$ into a perfect square, a squared parenthesis. Add $25$ to both sides to keep the same circle:
$(x^2-10x+25)+(y^2+4y)=20+25.$
Step 2
Complete the y-square
Half of the $y$ -term's coefficient, $4$ , is $2$ ; its square is $4$ . Add $4$ to both sides so $y^2+4y$ becomes a perfect square too:
$(x^2-10x+25)+(y^2+4y+4)=20+25+4.$
Step 3
Write the squared parentheses
The first group is $(x-5)^2$ , and the second is $(y+2)^2$ . Rewrite both groups:
$(x-5)^2+(y+2)^2=20+25+4.$
Step 4
Finish the right side
Type $20+25+4$ in Desmos; it prints $49$ . So the same circle in standard form is
$(x-5)^2+(y+2)^2=49.$
Choice A.

Question 788·Medium·Math

Strategy

Hints

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Step-by-step

Lessons that teach this