Answer:
[tex]radius = \sqrt{13}[/tex] or [tex]radius = 3.61[/tex]
Step-by-step explanation:
Given
Points:
A(-3,2) and B(-2,3)
Required
Determine the radius of the circle
First, we have to determine the center of the circle;
Since the circle has its center on the x axis; the coordinates of the center is;
[tex]Center = (x,0)[/tex]
Next is to determine the value of x through the formula of radius;
[tex]radius = \sqrt{(x_1 - x)^2 + (y_1 - y)^2} = \sqrt{(x_2 - x)^2 + (y_2 - y)^2}[/tex]
Considering the given points
[tex]A(x_1,y_1) = A(-3,2)[/tex]
[tex]B(x_2,y_2) = B(-2,3)[/tex]
[tex]Center(x,y) =Center (x,0)[/tex]
Substitute values for [tex]x,y,x_1,y_1,x_2,y_2[/tex] in the above formula
We have:
[tex]\sqrt{(-3 - x)^2 + (2 - 0)^2} = \sqrt{(-2 - x)^2 + (3 - 0)^2}[/tex]
Evaluate the brackets
[tex]\sqrt{(-(3 + x))^2 + 2^2} = \sqrt{(-(2 + x))^2 + 3 ^2}[/tex]
[tex]\sqrt{(-(3 + x))^2 + 4} = \sqrt{(-(2 + x))^2 + 9}[/tex]
Eva;uate all squares
[tex]\sqrt{(-(3 + x))(-(3 + x)) + 4} = \sqrt{(-(2 + x))(-(2 + x)) + 9}[/tex]
[tex]\sqrt{(3 + x)(3 + x) + 4} = \sqrt{(2 + x)(2 + x) + 9}[/tex]
Take square of both sides
[tex](3 + x)(3 + x) + 4 = (2 + x)(2 + x) + 9[/tex]
Evaluate the brackets
[tex]3(3 + x) +x(3 + x) + 4 = 2(2 + x) +x(2 + x) + 9[/tex]
[tex]9 + 3x +3x + x^2 + 4 = 4 + 2x +2x + x^2 + 9[/tex]
[tex]9 + 6x + x^2 + 4 = 4 + 4x + x^2 + 9[/tex]
Collect Like Terms
[tex]6x -4x + x^2 -x^2 = 4 -4 + 9 - 9[/tex]
[tex]2x = 0[/tex]
Divide both sides by 2
[tex]x = 0[/tex]
This implies the the center of the circle is
[tex]Center = (x,0)[/tex]
Substitute 0 for x
[tex]Center = (0,0)[/tex]
Substitute 0 for x and y in any of the radius formula
[tex]radius = \sqrt{(x_1 - 0)^2 + (y_1 - 0)^2}[/tex]
[tex]radius = \sqrt{(x_1)^2 + (y_1)^2}[/tex]
Considering that we used x1 and y1;
In this case we have that; [tex]A(x_1,y_1) = A(-3,2)[/tex]
Substitute -3 for x1 and 2 for y1
[tex]radius = \sqrt{(-3)^2 + (2)^2}[/tex]
[tex]radius = \sqrt{13}[/tex]
[tex]radius = 3.61[/tex] ---Approximated
