Question

write the equation in standard form for the circle $x^{2}+y^{2}-2y - 39 = 0$.

Explanation:

Step1: Group x and y terms

The given equation is \(x^{2}+y^{2}-2y - 39=0\). Group the \(x\) terms and \(y\) terms: \(x^{2}+(y^{2}-2y)-39 = 0\).

Step2: Complete the square for y - terms

For the \(y\) - terms \(y^{2}-2y\), we use the formula \((a - b)^2=a^{2}-2ab + b^{2}\). Here, \(a = y\) and \(2ab=2y\), so \(b = 1\). Then \(y^{2}-2y=(y - 1)^{2}-1\).
Substitute this into the equation: \(x^{2}+(y - 1)^{2}-1-39=0\).

Step3: Simplify the equation

Simplify the constants: \(x^{2}+(y - 1)^{2}-40 = 0\), then \(x^{2}+(y - 1)^{2}=40\).

Answer:

\(x^{2}+(y - 1)^{2}=40\)