Question

write the equation of the circle centered at $(-6,8)$ that passes through $(0,-3)$.

Explanation:

Step1: Find the radius

The radius \(r\) of the circle is the distance between the center \((-6,8)\) and the point \((0,-3)\) on the circle.
Use the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
Here \(x_1=-6,y_1 = 8,x_2=0,y_2=-3\).
\(r=\sqrt{(0-(-6))^2+(-3 - 8)^2}=\sqrt{6^2+(-11)^2}=\sqrt{36 + 121}=\sqrt{157}\)

Step2: Write the equation of the circle

The standard form of the equation of a circle is \((x - h)^2+(y - k)^2=r^2\), where \((h,k)\) is the center of the circle.
Given \(h=-6,k = 8,r^2 = 157\)
The equation is \((x+6)^2+(y - 8)^2=157\)

Answer:

\((x + 6)^2+(y-8)^2=157\)