Question

the equation for a circle is shown below.

$x^{2}+y^{2}-14x + 10y + 25 = 0$

what is the standard form of this equation?

$(x - 14)^{2}+(y + 10)^{2}=49$

$(x - 7)^{2}+(y + 5)^{2}=49$

$(x - 14)^{2}+(y + 10)^{2}=7$

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

Explanation:

Step1: Group \(x\) and \(y\) terms

$$x^{2}-14x + y^{2}+10y+25 = 0$$

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

For \(x^{2}-14x\), \((\frac{-14}{2})^{2}=49\). So \(x^{2}-14x=(x - 7)^{2}-49\)

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

For \(y^{2}+10y\), \((\frac{10}{2})^{2}=25\). So \(y^{2}+10y=(y + 5)^{2}-25\)

Step4: Substitute back into the equation

\((x - 7)^{2}-49+(y + 5)^{2}-25 + 25=0\)
Simplify: \((x - 7)^{2}+(y + 5)^{2}=49\)

Answer:

\((x - 7)^{2}+(y + 5)^{2}=49\) (the second option)