Question

rewrite the equation of the circle $x^{2}+y^{2}+4x - 6y - 3 = 0$ into standard form. enter your answers in the boxes to complete the equation. be sure to add an addition or subtraction symbol where needed. $(xsquare)^{2}+(ysquare)^{2}=square$

Explanation:

Step1: Group x - terms and y - terms

$(x^{2}+4x)+(y^{2}-6y)-3 = 0$

Step2: Complete the square for x - terms

For the $x^{2}+4x$ part, using the formula $(a + b)^2=a^{2}+2ab + b^{2}$, where $a = x$ and $2b = 4$ (so $b = 2$), we add $2^{2}=4$ to both sides of the equation.
$(x^{2}+4x + 4)+(y^{2}-6y)-3=4$

Step3: Complete the square for y - terms

For the $y^{2}-6y$ part, using the formula $(a - b)^2=a^{2}-2ab + b^{2}$, where $a = y$ and $2b = 6$ (so $b = 3$), we add $3^{2}=9$ to both sides of the equation.
$(x^{2}+4x + 4)+(y^{2}-6y+9)-3=4 + 9$

Step4: Rewrite in standard form

The standard form of a circle equation is $(x - h)^{2}+(y - k)^{2}=r^{2}$, where $(h,k)$ is the center of the circle and $r$ is the radius.
$(x + 2)^{2}+(y - 3)^{2}=16$

Answer:

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