Question

which is the equation for a circle with center at (-2, -4) that passes through the point (3, 8)?
$(x - 2)^2 + (y + 4)^2 = 169$
$(x + 2)^2 + (y + 4)^2 = 144$
$(x - 2)^2 + (y + 4)^2 = 144$
$(x + 2)^2 + (y + 4)^2 = 169$

Explanation:

Step1: Recall circle equation formula

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. Here, \(h=-2\), \(k = - 4\), so the equation starts as \((x+2)^2+(y + 4)^2=r^2\).

Step2: Calculate the radius squared

The circle passes through \((3,8)\). The distance between \((-2,-4)\) and \((3,8)\) is the radius. Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), so \(r^2=(3+2)^2+(8 + 4)^2=5^2+12^2 = 25 + 144=169\).

Step3: Form the circle equation

Substitute \(h=-2\), \(k=-4\), and \(r^2 = 169\) into the standard equation: \((x + 2)^2+(y + 4)^2=169\).

Answer:

\((x + 2)^2+(y + 4)^2=169\) (the fourth option: \((x + 2)^{2}+(y + 4)^{2}=169\))