Question

complete the square and write the given equation in standard form. then give the center and radius of the circle and graph the equation.
x² + y² + 6x - 2y - 26 = 0
the equation of the circle in standard form is .
(simplify your answer.)

Explanation:

Step1: Group x - terms and y - terms

$(x^{2}+6x)+(y^{2}-2y)=26$

Step2: Complete the square for x - terms

For the expression $x^{2}+6x$, we take half of the coefficient of $x$ (which is $6$), square it. Half of $6$ is $3$, and $3^{2}=9$. Add 9 to both sides of the equation:
$(x^{2}+6x + 9)+(y^{2}-2y)=26 + 9$

Step3: Complete the square for y - terms

For the expression $y^{2}-2y$, half of the coefficient of $y$ (which is $-2$) is $-1$, and $(-1)^{2}=1$. Add 1 to both sides of the equation:
$(x^{2}+6x + 9)+(y^{2}-2y+1)=26 + 9+1$

Step4: Rewrite in standard form

The standard form of a circle equation is $(x - a)^{2}+(y - b)^{2}=r^{2}$, where $(a,b)$ is the center and $r$ is the radius.
$(x + 3)^{2}+(y - 1)^{2}=36$

Answer:

$(x + 3)^{2}+(y - 1)^{2}=36$