Question

the equation of a circle in general form is ( x^{2}+y^{2}+22x + 14y-55 = 0 ). what is the equation of the circle in standard form? ( (x + 11)^{2}+(y + 7)^{2}=121 ) ( (x + 22)^{2}+(y + 14)^{2}=121 ) ( (x + 11)^{2}+(y + 7)^{2}=225 ) ( (x + 22)^{2}+(y + 14)^{2}=225 )

Explanation:

Step1: Complete the square for \(x\) terms

Given \(x^{2}+22x\), we use the formula \((a + b)^{2}=a^{2}+2ab + b^{2}\). For \(x^{2}+22x\), if \(a = x\) and \(2ab=22x\), then \(b = 11\). So \(x^{2}+22x=(x + 11)^{2}-121\)

Step2: Complete the square for \(y\) terms

Given \(y^{2}+14y\), using \((a + b)^{2}=a^{2}+2ab + b^{2}\). If \(a=y\) and \(2ab = 14y\), then \(b = 7\). So \(y^{2}+14y=(y + 7)^{2}-49\)

Step3: Rewrite the general equation in standard form

The general equation \(x^{2}+y^{2}+22x + 14y-55 = 0\) can be rewritten as \((x^{2}+22x)+(y^{2}+14y)=55\). Substituting the completed - square forms: \((x + 11)^{2}-121+(y + 7)^{2}-49=55\). Then \((x + 11)^{2}+(y + 7)^{2}=55 + 121+49\). Calculate \(55+121 + 49=225\)

Answer:

\((x + 11)^{2}+(y + 7)^{2}=225\)