Question

in the graph on the right, a line segment through the center of the circle intersects the circle at the points (1,2) and (11, 10) 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.)

Explanation:

Step1: Find center coordinates

The center of the circle is the mid - point of the line segment through the center (diameter endpoints). The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. Here, $x_1=1,y_1 = 2,x_2=11,y_2 = 10$. So, the x - coordinate of the center is $\frac{1+11}{2}=\frac{12}{2}=6$, and the y - coordinate of the center is $\frac{2 + 10}{2}=\frac{12}{2}=6$.

Step2: Find the radius

The radius is the distance from the center $(6,6)$ to either of the given points, say $(1,2)$. The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Substitute $x_1 = 6,y_1=6,x_2 = 1,y_2 = 2$: $r=\sqrt{(1 - 6)^2+(2 - 6)^2}=\sqrt{(-5)^2+(-4)^2}=\sqrt{25 + 16}=\sqrt{41}$.

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, $h = 6,k = 6,r=\sqrt{41}$, so the equation is $(x - 6)^2+(y - 6)^2=41$.

Answer:

a. $(6,6)$
b. $\sqrt{41}$
c. $(x - 6)^2+(y - 6)^2=41$