Question

write the equation in standard form for the circle $x^{2}+y^{2}-8x - 9 = 0$.

Explanation:

Step1: Group x - terms and y - terms

Given the equation \(x^{2}+y^{2}-8x - 9=0\). Group the \(x\) - terms together: \((x^{2}-8x)+y^{2}=9\).

Step2: Complete the square for the x - terms

For the quadratic expression \(x^{2}-8x\), using the formula \((a - b)^2=a^{2}-2ab + b^{2}\). Here \(a = x\) and \(2ab=8x\), so \(b = 4\). Then \(x^{2}-8x=(x - 4)^{2}-16\).
The equation becomes \((x - 4)^{2}-16+y^{2}=9\).

Step3: Simplify to get the standard form

Add 16 to both sides of the equation: \((x - 4)^{2}+y^{2}=9 + 16\).

Answer:

\((x - 4)^{2}+y^{2}=25\)