Question

enter possible values to show an equation of a circle where the center of the circle lies in the second quadrant of the coordinate plane and the radius is twice the value of the center’s y-coordinate.
$(x - \square)^2 + (y - \square)^2 = \square$

Explanation:

Step1: Recall circle equation

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

Step2: Determine center coordinates

In the second quadrant, \(h<0\) and \(k>0\). Let's choose \(h = - 2\) (negative) and \(k = 3\) (positive).

Step3: Calculate radius

Radius \(r = 2k\). Substituting \(k = 3\), we get \(r=2\times3 = 6\).

Step4: Substitute into equation

Substitute \(h=-2\), \(k = 3\), and \(r = 6\) into the circle equation: \((x - (-2))^2+(y - 3)^2=6^2\), which simplifies to \((x + 2)^2+(y - 3)^2 = 36\).

Answer:

\((x - (-2))^2+(y - 3)^2 = 36\) (or with \(h=-2\), \(k = 3\), \(r^2 = 36\) filled in the boxes as \(-2\), \(3\), \(36\) respectively)