Question

in the xy-plane, the graph of $(2x^{2}+14x)+2y^{2}-8y = 17.5$ is a circle. what is the centerpoint of the circle?
(a) $(-7,-4)$
(b) $(-3.5,2)$
(c) $(3.5,-2)$
(d) $(5,5)$

Explanation:

Step1: Divide the equation by 2

$$\begin{align*} \frac{2x^{2}+14x + 2y^{2}-8y}{2}&=\frac{17.5}{2}\\ x^{2}+7x+y^{2}-4y& = 8.75 \end{align*}$$

Step2: Complete the square for \(x\) - terms

For \(x^{2}+7x\), using the formula \((a + b)^2=a^{2}+2ab + b^{2}\), here \(a=x\), \(2ab = 7x\), so \(b=\frac{7}{2}\). Then \(x^{2}+7x=(x + \frac{7}{2})^{2}-\frac{49}{4}\)

Step3: Complete the square for \(y\) - terms

For \(y^{2}-4y\), using the formula \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a = y\), \(2ab=4y\), so \(b = 2\). Then \(y^{2}-4y=(y - 2)^{2}-4\)

Step4: Substitute back into the equation

$$\begin{align*} (x+\frac{7}{2})^{2}-\frac{49}{4}+(y - 2)^{2}-4&=8.75\\ (x+\frac{7}{2})^{2}+(y - 2)^{2}&=8.75+\frac{49}{4}+4\\ (x + 3.5)^{2}+(y - 2)^{2}&=8.75+12.25 + 4\\ (x + 3.5)^{2}+(y - 2)^{2}&=25 \end{align*}$$

The standard form of a circle equation is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center of the circle.

Answer:

B. \((-3.5,2)\)