SOLUTION: Find the equation for the circle with a diameter whose endpoints are​ (1,3) and (4,3)

Algebra ->  Circles -> SOLUTION: Find the equation for the circle with a diameter whose endpoints are​ (1,3) and (4,3)      Log On


   

Question 1151541: Find the equation for the circle with a diameter whose endpoints are​ (1,3) and (4,3)
Found 2 solutions by Alan3354, MathLover1:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Find the equation for the circle with a diameter whose endpoints are​ (1,3) and (4,3)
--------------
The diameter = 3 obviously.
The center is the midpoint (2.5,3)
r = 1.5
--------------
%28x+-+2.5%29%5E2+%2B+%28y+-+3%29%5E2+=+1.5%5E2

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
circle:
%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2
if a diameter whose endpoints are​ (1,3) and (4,3), the length is the distance between them
d=sqrt%284-1%29%5E2%2B%283-3%29%5E2%29
d=sqrt%283%5E2%29
d=3 => r=3%2F2
so far
%28x-h%29%5E2%2B%28y-k%29%5E2=%283%2F2%29%5E2

use given points to find h and k

%281-h%29%5E2%2B%283-k%29%5E2=9%2F4
h%5E2+-+2+h+%2B+k%5E2+-+6+k+%2B+10+=+9%2F4.......eq.1

%284-h%29%5E2%2B%283-k%29%5E2=9%2F4
h%5E2+-+8+h+%2B+k%5E2+-+6+k+%2B+25+=+9%2F4.......eq.2

subtract eq.2 from eq.1

h%5E2+-+2+h+%2B+k%5E2+-+6+k+%2B+10-%28h%5E2+-+8+h+%2B+k%5E2+-+6+k+%2B+25+%29+=+9%2F4-9%2F4
h%5E2+-+2+h+%2B+k%5E2+-+6+k+%2B+10-h%5E2+%2B8+h+-+k%5E2+%2B+6+k+-+25++=+0
+++10+%2B6+h++-+25++=+0
+++6+h++-+15++=+0
++++h++=+15%2F6++
++++h++=+5%2F2++

find k

%285%2F2%29%5E2+-+2+%285%2F2%29+%2B+k%5E2+-+6+k+%2B+10+=+9%2F4.......eq.1
25%2F4+-+5+%2B+k%5E2+-+6+k+%2B+10+=+9%2F4
++k%5E2+-+6+k++=+9%2F4-25%2F4%2B5-10
++k%5E2+-+6+k++=+-9
++k%5E2+-+6+k++%2B9=0
++%28k+-+3%29%5E2=0=>k=3

and, your equation of circle is:
%28x-5%2F2%29%5E2%2B%28y-3%29%5E2=%283%2F2%29%5E2