Question

put the equation in standard form.
$x^2 + y^2 - 18x + 14y + 9 = 0$
$(x - ?)^2 + (y + \square)^2 = \square$

Explanation:

Step1: Group x and y terms

Group the \(x\)-terms and \(y\)-terms together: \((x^2 - 18x) + (y^2 + 14y) = -9\)

Step2: Complete the square for x

For \(x^2 - 18x\), take half of \(-18\) (which is \(-9\)), square it: \((-9)^2 = 81\). Add 81 to both sides.
For \(y^2 + 14y\), take half of \(14\) (which is \(7\)), square it: \(7^2 = 49\). Add 49 to both sides.
So we have: \((x^2 - 18x + 81) + (y^2 + 14y + 49) = -9 + 81 + 49\)

Step3: Factor and simplify

Factor the perfect squares: \((x - 9)^2 + (y + 7)^2 = 121\)

Answer:

\((x - \boldsymbol{9})^2 + (y + \boldsymbol{7})^2 = \boldsymbol{121}\)