Question

right triangle abc is located at a(-1,-2), b(-1,1), and c(3,1) on a coordinate plane. what is the equation of a circle a with radius \overline{ac}?
(x - 3)^2 + (y - 1)^2 = 25
(x - 3)^2 + (y - 1)^2 = 16
(x + 1)^2 + (y + 2)^2 = 25
(x + 1)^2 + (y + 2)^2 = 9

Explanation:

Step1: Recall the standard form of a circle equation

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

Step2: Identify the center and radius

The center of circle \(A\) is \(A(-1,-2)\), so \(h=-1\) and \(k = - 2\). The radius \(r = AC\). Calculate the distance \(AC\) using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). Here \(x_1=-1,y_1=-2,x_2 = 3,y_2=1\). Then \(AC=\sqrt{(3+1)^2+(1 + 2)^2}=\sqrt{16 + 9}=\sqrt{25}=5\), so \(r = 5\) and \(r^2=25\).

Step3: Substitute into the circle equation

Substitute \(h=-1,k=-2,r^2 = 25\) into \((x - h)^2+(y - k)^2=r^2\). We get \((x+1)^2+(y + 2)^2=25\).

Answer:

\((x + 1)^2+(y+2)^2=25\)