Question

write the equation in standard form for the circle ( 3 x ^ { 2 } + 3 y ^ { 2 } - 30 x + 42 y + 87 = 0 ).

Explanation:

Step1: Divide the equation by 3

$$\begin{align*} \frac{3x^{2}+3y^{2}-30x + 42y+87}{3}&=\frac{0}{3}\\ x^{2}+y^{2}-10x + 14y+29&=0 \end{align*}$$

Step2: Group the x - terms and y - terms

$$(x^{2}-10x)+(y^{2}+14y)+29 = 0$$

Step3: Complete the square for the x - terms

For \(x^{2}-10x\), using the formula \((a - b)^2=a^{2}-2ab + b^{2}\), where \(a = x\) and \(2ab=10x\), so \(b = 5\). Then \(x^{2}-10x=(x - 5)^{2}-25\)

Step4: Complete the square for the y - terms

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

Step5: Substitute back into the equation

$$\begin{align*} (x - 5)^{2}-25+(y + 7)^{2}-49+29&=0\\ (x - 5)^{2}+(y + 7)^{2}-45&=0\\ (x - 5)^{2}+(y + 7)^{2}&=45 \end{align*}$$

Answer:

\((x - 5)^{2}+(y + 7)^{2}=45\)