Question

question 5 of 10
the circle given by $x^{2}+y^{2}-2y - 11=0$ can be written in standard form like this: $x^{2}+(y - k)^{2}=12$.
what is the value of $k$ in this equation?

Explanation:

Step1: Complete the square for y - terms

Given $x^{2}+y^{2}-2y - 11=0$. Focus on $y^{2}-2y$. Using the formula $(a - b)^2=a^{2}-2ab + b^{2}$, for $y^{2}-2y$, we have $a = y$ and $2ab=2y$, so $b = 1$. Then $y^{2}-2y=(y - 1)^{2}-1$. The equation becomes $x^{2}+(y - 1)^{2}-1-11=0$.

Step2: Simplify the equation

$x^{2}+(y - 1)^{2}-12=0$, which can be rewritten as $x^{2}+(y - 1)^{2}=12$.

Answer:

$1$