Question

question 13 (multiple choice worth 1 points) (07.03 hc) right triangle abc is located at a(-1, -2), b (-1, 1), and c (-5, 1) on a coordinate plane. what is the equation of a circle a with radius \\(\overline{ac}\\)? \\(\bigcirc (x + 1)^2 + (y + 2)^2 = 9\\) \\(\bigcirc (x + 5)^2 + (y - 1)^2 = 16\\) \\(\bigcirc (x + 1)^2 + (y + 2)^2 = 25\\) \\(\bigcirc (x + 5)^2 + (y - 1)^2 = 25\\)

Explanation:

Step1: Identify the center of the circle

The circle is centered at point \( A(-1, -2) \). The standard form of a circle's equation is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \( r \) is the radius. So here, \( h = -1 \) and \( k = -2 \). Substituting these into the standard form, we get \((x - (-1))^2 + (y - (-2))^2 = r^2\), which simplifies to \((x + 1)^2 + (y + 2)^2 = r^2\).

Step2: Calculate the radius (length of \( AC \))

Points \( A(-1, -2) \) and \( C(-5, 1) \). Use the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
Substitute \( x_1 = -1, y_1 = -2, x_2 = -5, y_2 = 1 \):

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So the radius \( r = 5 \), and \( r^2 = 25 \).

Step3: Write the equation of the circle

Substitute \( h = -1, k = -2, r^2 = 25 \) into the standard form: \((x + 1)^2 + (y + 2)^2 = 25\).

Answer:

\((x + 1)^2 + (y + 2)^2 = 25\) (the second option: \((x + 1)^2 + (y + 2)^2 = 25\))