Question

the equation of a circle x^2 + y^2 + 6y=7. What are the coordinates of the center and the length of the radius of the circle?

A) center (0,3) and radius 4
B) center (0,-3) and radius 4
C) center (0,3) and radius 16
D) center (0,-3) and radius 16

Answer 1

So the first step when dealing with circle problems is to put the equation in standard form.

x² + y² + 6y = 7

Original Problem.

(x+0)² + (y² + 6y) = 7

Expanded x variable and put y values in parentheses.

(x+0)² + (y² + 6y + 9) = 7 + 9

Added 9 to both sides.

(x+0)² + (y+3)² = 16

Factored out y variables.

(x+0)² + (y+3)² = 4²

Standard form.

Now to find the center, just look inside the parentheses. Inside the x set there is the x value, inside the y set there is the y value. (x+0) and (y+3). The center is (0,3). That eliminates answer choices B) and D). Now the radius is the square root of the last number in the standard equation. 4². The square root of 4 squared is 4. Therefore the radius is 4. That eliminates answer C).

That leaves us with one answer: A) center (0,3) and radius 4.

Hope that helps!

Answer 2

Answer:

Option B) center $(0,-3)$ and radius $4$

Step-by-step explanation:

we know that

The equation of the circle into center radius form is equal to

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

where

(h,k) is the center of the circle

r is the radius of the circle

In this problem we have

$x^{2} +y^{2}+6y=7$

so

convert to center radius form

Complete the square. Remember to balance the equation by adding the same constants to each side

$x^{2} +(y^{2}+6y+9)=7+9$

$x^{2} +(y^{2}+6y+9)=16$

Rewrite as perfect squares

$x^{2} +(y+3)^{2}=16$

$x^{2} +(y+3)^{2}=4^{2}$

The center is the point $(0,-3)$

The radius is $4$ units