Question

write the equation of the circle centered at (0,2) that passes through (-10,-8).

Explanation:

Step1: Recall the standard equation of a circle

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

Step2: Determine the center of the circle

Given that the center of the circle is \((h,k)=(0,2)\), so the equation becomes \((x - 0)^2+(y - 2)^2=r^2\), or \(x^{2}+(y - 2)^2=r^2\).

Step3: Calculate the radius

Since the circle passes through the point \((- 10,-8)\), substitute \(x=-10\) and \(y = - 8\) into the equation \(x^{2}+(y - 2)^2=r^2\).

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Answer:

The equation of the circle is \(x^{2}+(y - 2)^2=200\)