Question

HELP ASAP

The points $(-3,2)$ and $(-2,3)$ lie on a circle whose center is on the $x$-axis. What is the radius of the circle?

Answer 1

`radius = √(13)` or
`radius = 3.61`

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 = √((x_1 - x)^2 + (y_1 - y)^2) = √((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:

`√((-3 - x)^2 + (2 - 0)^2) = √((-2 - x)^2 + (3 - 0)^2)`

Evaluate the brackets

`√((-(3 + x))^2 + 2^2) = √((-(2 + x))^2 + 3 ^2)`

`√((-(3 + x))^2 + 4) = √((-(2 + x))^2 + 9)`

Eva;uate all squares

`√((-(3 + x))(-(3 + x)) + 4) = √((-(2 + x))(-(2 + x)) + 9)`

`√((3 + x)(3 + x) + 4) = √((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 = √((x_1 - 0)^2 + (y_1 - 0)^2)`

`radius = √((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 = √((-3)^2 + (2)^2)`

`radius = √(13)`

`radius = 3.61` ---Approximated