Question

How do find the standard equation of the circle with endpoints of a diameter (3,6) and (-1,4)?

Answer 1

See a solution process below:

Explanation:

The equation for a circle is:

`(x - color(red)(a))^2 + (y - color(red)(b))^2 = color(blue)(r)^2`

Where `(color(red)(a), color(red)(b))` is the center of the circle and `color(blue)(r)` is the radius of the circle.

First, we can find the center point of the circle. Because the two points from the problem are the endpoints of a diameter, the midpoint of the line segment is the center point of the circle.

The formula to find the mid-point of a line segment give the two end points is:

`M = ((color(red)(x_1) + color(blue)(x_2))/2 , (color(red)(y_1) + color(blue)(y_2))/2)`

Where `M` is the midpoint and the given points are:

`(color(red)(x_1), color(red)(y_1))` and `(color(blue)(x_2), color(blue)(y_2))`

Substituting the values from the two points in the problem gives:

`M = ((color(red)(3) + color(blue)(-1))/2 , (color(red)(6) + color(blue)(4))/2)`

`M = ((color(red)(3) - color(blue)(1))/2 , (color(red)(6) + color(blue)(4))/2)`

`M = (2/2 , 10/2)`

`M = (1 , 5)`

The Center Point of the Circle is: `(color(red)(1), color(red)(5))`

Next, we need to find the radius of the circle. We can use the formula for distance to find the distance between one of the points from the problem and the center point of the circle.

Substituting the values from the problem gives: The formula for calculating the distance between two points is:

`d = sqrt((color(red)(x_2) - color(blue)(x_1))^2 + (color(red)(y_2) - color(blue)(y_1))^2)`

Substituting the values from the first point in the problem and the center point gives:

`d = sqrt((color(red)(3) - color(blue)(1))^2 + (color(red)(6) - color(blue)(5))^2)`

`d = sqrt(2^2 + 1^2)`

`d = sqrt(4 + 1)`

`d = sqrt(5)`

The Radius of the Circle is: `color(blue)(sqrt(5))`

Substituting the center point and radius we calculated into the formula for a circle gives:

`(x - color(red)(1))^2 + (y - color(red)(5))^2 = (color(blue)(sqrt(5)))^2`

`(x - color(red)(1))^2 + (y - color(red)(5))^2 = 5`