Question

put the equation in standard form.

$x^{2}+y^{2}-18x + 14y + 9 = 0$

$(x - ?)^{2}+(y + )^{2}= $

Explanation:

Step1: Group x and y terms

$$(x^{2}-18x)+(y^{2}+14y)+9 = 0$$

Step2: Complete the square for x - terms

For \(x^{2}-18x\), using the formula \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a = x\) and \(2b=18\) (so \(b = 9\)). Then \(x^{2}-18x=(x - 9)^{2}-81\)

Step3: Complete the square for y - terms

For \(y^{2}+14y\), using the formula \((a + b)^2=a^{2}+2ab + b^{2}\), here \(a = y\) and \(2b = 14\) (so \(b=7\)). Then \(y^{2}+14y=(y + 7)^{2}-49\)

Step4: Substitute back into the equation

\((x - 9)^{2}-81+(y + 7)^{2}-49+9=0\)
\((x - 9)^{2}+(y + 7)^{2}-81-49 + 9=0\)
\((x - 9)^{2}+(y + 7)^{2}-121=0\)
\((x - 9)^{2}+(y + 7)^{2}=121\)

Answer:

\((x - 9)^{2}+(y + 7)^{2}=121\), so the first box is \(9\), the second box is \(7\), and the third box is \(121\)