Question

complete the square and write the equation of the circle in standard form. then find the center and radius of the circle and graph the equation.
$x^{2}+y^{2}+7x - 2y-1 = 0$

Explanation:

Step1: Group x - terms and y - terms

$(x^{2}+7x)+(y^{2}-2y)=1$

Step2: Complete the square for x - terms

For the $x^{2}+7x$ part, take half of the coefficient of $x$ ($\frac{7}{2}$), square it ($\frac{49}{4}$) and add it to both sides of the equation.
$(x^{2}+7x+\frac{49}{4})+(y^{2}-2y)=1 + \frac{49}{4}$

Step3: Complete the square for y - terms

For the $y^{2}-2y$ part, take half of the coefficient of $y$ (- 1), square it (1) and add it to both sides of the equation.
$(x^{2}+7x+\frac{49}{4})+(y^{2}-2y + 1)=1+\frac{49}{4}+1$

Step4: Write in standard form

The standard form of the equation of a circle is $(x - a)^{2}+(y - b)^{2}=r^{2}$, where $(a,b)$ is the center and $r$ is the radius.
$(x+\frac{7}{2})^{2}+(y - 1)^{2}=\frac{4 + 49+4}{4}=\frac{57}{4}$

Step5: Find the center and radius

Comparing with the standard form, the center of the circle is $(-\frac{7}{2},1)$ and the radius $r=\sqrt{\frac{57}{4}}=\frac{\sqrt{57}}{2}$

Answer:

The standard - form of the equation is $(x+\frac{7}{2})^{2}+(y - 1)^{2}=\frac{57}{4}$, the center is $(-\frac{7}{2},1)$ and the radius is $\frac{\sqrt{57}}{2}$