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

1 Answer
Mar 23, 2018

(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:
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