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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