Question

Mrs. Culland is finding the center of a circle whose equation

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