Question

HELP ASAP

Answer 1

Answer:

$radius = \sqrt{13}$ or $radius = 3.61$

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;

$Center = (x,0)$

Next is to determine the value of x through the formula of radius;

$radius = \sqrt{(x_1 - x)^2 + (y_1 - y)^2} = \sqrt{(x_2 - x)^2 + (y_2 - y)^2}$

Considering the given points

$A(x_1,y_1) = A(-3,2)$

$B(x_2,y_2) = B(-2,3)$

$Center(x,y) =Center (x,0)$

Substitute values for $x,y,x_1,y_1,x_2,y_2$ in the above formula

We have:

$\sqrt{(-3 - x)^2 + (2 - 0)^2} = \sqrt{(-2 - x)^2 + (3 - 0)^2}$

Evaluate the brackets

$\sqrt{(-(3 + x))^2 + 2^2} = \sqrt{(-(2 + x))^2 + 3 ^2}$

$\sqrt{(-(3 + x))^2 + 4} = \sqrt{(-(2 + x))^2 + 9}$

Eva;uate all squares

$\sqrt{(-(3 + x))(-(3 + x)) + 4} = \sqrt{(-(2 + x))(-(2 + x)) + 9}$

$\sqrt{(3 + x)(3 + x) + 4} = \sqrt{(2 + x)(2 + x) + 9}$

Take square of both sides

$(3 + x)(3 + x) + 4 = (2 + x)(2 + x) + 9$

Evaluate the brackets

$3(3 + x) +x(3 + x) + 4 = 2(2 + x) +x(2 + x) + 9$

$9 + 3x +3x + x^2 + 4 = 4 + 2x +2x + x^2 + 9$

$9 + 6x + x^2 + 4 = 4 + 4x + x^2 + 9$

Collect Like Terms

$6x -4x + x^2 -x^2 = 4 -4 + 9 - 9$

$2x = 0$

Divide both sides by 2

$x = 0$

This implies the the center of the circle is

$Center = (x,0)$

Substitute 0 for x

$Center = (0,0)$

Substitute 0 for x and y in any of the radius formula

$radius = \sqrt{(x_1 - 0)^2 + (y_1 - 0)^2}$

$radius = \sqrt{(x_1)^2 + (y_1)^2}$

Considering that we used x1 and y1;

In this case we have that; $A(x_1,y_1) = A(-3,2)$

Substitute -3 for x1 and 2 for y1

$radius = \sqrt{(-3)^2 + (2)^2}$

$radius = \sqrt{13}$

$radius = 3.61$ ---Approximated