Question

in the graph on the right, a line segment through the center of the circle intersects the circle at the points (3,1) and (7,13) as shown.
a. find the coordinates of the circles center.
b. find the radius of the circle.
c. use your answers from parts (a) and (b) to write the standard form of the circles equation.
the center is (type an ordered pair.)
the radius is
(simplify your answer. type an exact answer, using radicals as needed.)
the equation for the circle in standard form is
(simplify your answer.)

Explanation:

Step1: Find the center of the circle

The center of the circle is the mid - point of the line segment joining the points \((3,1)\) and \((7,13)\).
The mid - point formula is \((\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})\).
Here \(x_1 = 3,y_1 = 1,x_2 = 7,y_2 = 13\).
\(x=\frac{3 + 7}{2}=\frac{10}{2}=5\)
\(y=\frac{1+13}{2}=\frac{14}{2}=7\)

Step2: Find the radius of the circle

The radius is the distance from the center \((5,7)\) to either of the points \((3,1)\) or \((7,13)\).
Use the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
Take \((x_1,y_1)=(5,7)\) and \((x_2,y_2)=(3,1)\)
\(r=\sqrt{(3 - 5)^2+(1 - 7)^2}=\sqrt{(-2)^2+(-6)^2}=\sqrt{4 + 36}=\sqrt{40}=2\sqrt{10}\)

Step3: Write the standard form of the circle's equation

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.
Here \(h = 5,k = 7,r = 2\sqrt{10}\)
\((x - 5)^2+(y - 7)^2=(2\sqrt{10})^2\)
\((x - 5)^2+(y - 7)^2 = 40\)

Answer:

• The center is \((5,7)\)
• The radius is \(2\sqrt{10}\)
• The equation for the circle in standard form is \((x - 5)^2+(y - 7)^2 = 40\)