Question

write the equation in standard form for the circle $x^2 + y^2 - 21 = 4y$.

Explanation:

Step1: Rearrange the equation

We start with the given equation \(x^{2}+y^{2}-21 = 4y\). First, we want to group the \(x\) terms and \(y\) terms together. So we rewrite the equation as \(x^{2}+y^{2}-4y=21\).

Step2: Complete the square for the y - terms

For the \(y\) - terms \(y^{2}-4y\), we use the formula for completing the square. The coefficient of \(y\) is \(- 4\), half of it is \(-2\), and squaring it gives \((-2)^{2}=4\). We add and subtract this value (but since we are dealing with an equation, we add it to both sides to keep the equation balanced).
So we have \(x^{2}+(y^{2}-4y + 4)=21 + 4\).

Step3: Write in standard form

The standard form of the equation of a circle is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center of the circle and \(r\) is the radius.
For the \(x\) - terms, \(x^{2}=(x - 0)^{2}\), and for the \(y\) - terms, \(y^{2}-4y + 4=(y - 2)^{2}\). And \(21+4 = 25\).
So the equation in standard form is \(x^{2}+(y - 2)^{2}=25\) or \((x - 0)^{2}+(y - 2)^{2}=5^{2}\).

Answer:

\(x^{2}+(y - 2)^{2}=25\)