Question

a circle has a radius of \\(\sqrt{131}\\) units and is centered at \\((0, -4.3)\\). write the equation of this circle.

Explanation:

Step1: Recall the standard form of a circle's 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: Identify the values of \(h\), \(k\), and \(r\)

Given the center \((0,-4.3)\), so \(h = 0\), \(k=-4.3\), and \(r=\sqrt{131}\).

Step3: Substitute the values into the standard form

Substitute \(h = 0\), \(k=-4.3\), and \(r=\sqrt{131}\) into \((x - h)^2+(y - k)^2=r^2\).
We get \((x - 0)^2+(y-(-4.3))^2=(\sqrt{131})^2\).
Simplify to \(x^{2}+(y + 4.3)^{2}=131\).

Answer:

\(x^{2}+(y + 4.3)^{2}=131\)