Question

question 3
what are the center and radius of the circle given by x² + y² + 22x + 18y - 14 = 0?
(11, 9); 196
(-11, -9); 6√6
(22, 18); 14
(11, 9); √14

Explanation:

Step1: Complete the square for x - terms

The general equation of a circle is $(x - a)^2+(y - b)^2=r^2$, where $(a,b)$ is the center and $r$ is the radius. For the $x$ - terms in $x^{2}+22x$, we have $x^{2}+22x=(x + 11)^{2}-121$.

Step2: Complete the square for y - terms

For the $y$ - terms in $y^{2}+18y$, we have $y^{2}+18y=(y + 9)^{2}-81$.

Step3: Rewrite the circle equation

The original equation $x^{2}+y^{2}+22x + 18y-14 = 0$ can be rewritten as $(x + 11)^{2}-121+(y + 9)^{2}-81-14 = 0$.

Step4: Simplify the equation

$(x + 11)^{2}+(y + 9)^{2}=121 + 81+14$.
$(x + 11)^{2}+(y + 9)^{2}=216$.

Step5: Identify the center and radius

The center of the circle is $(-11,-9)$ and the radius $r=\sqrt{216}=6\sqrt{6}$.

Answer:

B. (-11, -9); $6\sqrt{6}$