Question

question 4 | lesson 1 - 2 quiz
write the equation in standard form to find the center and radius of the circle.
$x^{2}+y^{2}+6x + 14y+37 = 0$
a) $(x - 3)^{2}+(y - 7)^{2}=21$; center $(3,7)$, radius $sqrt{21}$
b) $(x + 3)^{2}+(y + 7)^{2}=21$; center $(-3,-7)$, radius $21$
c) $(x + 3)^{2}+(y + 7)^{2}=21$; center $(-3,-7)$, radius $sqrt{21}$
d) $(x + 3)^{2}+(y + 7)^{2}=21$; center $(3,7)$, radius $21$

Explanation:

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

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

For \(x^{2}+6x\), using \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a=x\), \(2b = 6\Rightarrow b = 3\), so \(x^{2}+6x=(x + 3)^{2}-9\)

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

For \(y^{2}+14y\), using \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a=y\), \(2b=14\Rightarrow b = 7\), so \(y^{2}+14y=(y + 7)^{2}-49\)

Step4: Substitute back into the equation

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The standard form of a circle is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius. Here \(h=-3,k=-7,r=\sqrt{21}\)

Answer:

D. \((x + 3)^{2}+(y + 7)^{2}=21\); center \((-3,-7)\), radius \(\sqrt{21}\)