Question

the equation below describes a circle.
$x^2 + 14x + y^2 = 70$
what are the center and radius of the circle?

• the center is $(-7, 0)$ and the radius is $sqrt{119}$.
• the center is $(7, 0)$ and the radius is $sqrt{119}$.
• the center is $(7, 0)$ and the radius is $sqrt{70}$.
• the center is $(-7, 0)$ and the radius is $sqrt{70}$.

Explanation:

Step1: Recall the standard circle equation

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

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

Given the equation \(x^2 + 14x + y^2 = 70\). For the \(x\)-terms, take the coefficient of \(x\) (which is 14), divide by 2: \(14\div2 = 7\), then square it: \(7^2 = 49\). Add and subtract 49 (but we'll just add it to both sides to complete the square). So we rewrite the equation as:
\(x^2 + 14x + 49 + y^2 = 70 + 49\)

Step3: Rewrite in standard form

The left side becomes a perfect square: \((x + 7)^2 + y^2 = 119\). Which can be written as \((x - (-7))^2 + (y - 0)^2 = (\sqrt{119})^2\).

Step4: Identify center and radius

From the standard form, \(h = -7\), \(k = 0\), so the center is \((-7, 0)\), and \(r^2 = 119\), so \(r = \sqrt{119}\).

Answer:

The center is \((-7,0)\) and the radius is \(\sqrt{119}\). (The first option: The center is \((-7,0)\) and the radius is \(\sqrt{119}\).)