Question

Mrs. Culland is finding the center of a circle whose equation
is x2 + y2 + 6x + 4y – 3 = 0 by completing the square. Her
work is shown.
x2 + y2 + 6x + 4y - 3 = 0
x2 + 6x + y2 + 4y - 3 = 0
(x2 + 6x) + (12 + 4y) = 3
(x2 + 6x + 9) + (y2 + 4y + 4) = 3 + 9 + 4
Which completes the work correctly?
(x - 3)2 + ( - 2)= 42, so the center is (3,2).
(x + 3)2 + (y + 2)2 = 42, so the center is (3, 2).
(x - 3)2 + (x - 2)2 = 42, so the center is (-3,-2).
(x + 3)2 + (y + 2)2 = 42, so the center is (-3, -2).

Answer 1

We have been given that Mrs. Culland is finding the center of a circle whose equation  is $x^2+y^2+6x +4y-3 = 0$ by completing the square.

Her work is shown.

$x^2+6x+y^2 +4y-3 = 0$

$(x^2+6x)+(y^2 +4y)=3$

$(x^2+6x+9)+(y^2 +4y+4)=3+9+4$.

We can see that Mrs. Culland added half the square of coefficients of x and y terms on both sides to complete the square.

For x term: $(\frac{6}{2})^2=3^3=9$

For y term: $(\frac{2}{2})^2=2^3=4$

Our next step is to write the perfect squares in form $(a\pm b)^2$ as:

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

We know that standard equation of circle is $(x-h)^2+(y-k)^2=r^2$, where point (h,k) is center of circle and r is radius of circle.

We can rewrite our equation as:

$(x-(-3))^2+(y-(-2))^2=4^2$

Therefore, the center of circle is $(-3,-2)$ and equation of circle would be $(x+3)^2+(y+2)^2=4^2$.

Answer 2

Answer:

d ( (x + 3)2 + (y + 2)2 = 42, so the center is (–3, –2) )

Step-by-step explanation:

hope this helps :)