Question

What is an equation of a circle if the endpoints of one of its diameters are (2,3) and (4,9)?

Answer 1

`(x-3)^2+(y-6)^2=10`

Explanation:

Since the two given points, `(2, 3)` and `(4, 9)`, form a line segment, you can figure out:

1) The length of the segment (diameter of the circle), and
2) The midpoint of the segment (the center of the circle).

Let's start with the length of the segment:

Using the Pythagorean theorem (`a^2 + b^2 = c^2`), we can take the `Delta x` as our `a` value and `Delta y` as our `b` value. The length of the segment will be `c`.

`Delta x = x_2-x_1`
`Delta x = 4-2`
`Delta x = 2`
`a = 2`

`Delta y = y_2-y_1`
`Delta y = 9-3`
`Delta y = 6`
`b=6`

If we plug these in, we get:

`2^2+6^2=c^2`
`4+36=c^2`
`40=c^2`
`2sqrt(10)=c`

We have our diameter of the circle now, but we want the radius so we can write the equation of the circle later:

`2r=d`
`r=d/2`
`r=(2sqrt(10))/2`

`r=sqrt(10)`

We'll use this later.

Now, for the midpoint, which is much simpler:

We're going to find the middle of this line, and the center's coordinate pair will be: `((x_1+x_2)/2,(y_1+y_2)/2)`

This means finding the average between the `x` values of each point and the `y` values of each point, which will look like:

`((2+4)/2,(3+9)/2)`
`(6/2, 12/2)`
`(3,6)`

There's the center! Now we have all the info we need, and we just need to plug it into the general form of the circle equation:

`(x-h)^2+(y-k)^2=r^2`
Where:
`h` is the `x`-coordinate of the center,
`k` is the `y`-coordinate of the center, and
`r` is the radius of the circle.

We know all of these, so we can just plug them in to get:

`(x-3)^2+(y-6)^2=(sqrt(10))^2`, which simplifies to:
`(x-3)^2+(y-6)^2=10`

The segment is in purple, the midpoint is in red, and the circle is in green: