SOLUTION: the line x+3y=20 intersect the circle x'2+y'2-6x-8y=0 at the points P and Q. Find the equation of the circle that has [PQ] as diameter.

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Question 968104: the line x+3y=20 intersect the circle x'2+y'2-6x-8y=0 at the points P and Q.
Find the equation of the circle that has [PQ] as diameter.
Found 2 solutions by Edwin McCravy, Theo:
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
 
First we solve the system of equations: 

system%28x%2B3y=20%2C+x%5E2%2By%5E2-6x-8y=0%29 

to find where they intersect: 

Solve the first equation for x 
  
x+3y = 20 
   x = 20-3y 

Substitute (20-3y) for x in the 2nd equation: 

x%5E2%2By%5E2-6x-8y=0 

%2820-3y%29%5E2%2By%5E2-6%2820-3y%29-8y=0 

400-120y%2B9y%5E2%2By%5E2-120%2B18y-8y=0 

10y%5E2-110y%2B280+=+0 

Divide thru by 10: 

y%5E2-11y%2B28+=+0 

%28y-7%29%28y-4%29=0 

y-7 = 0;  y-4 = 0 
  y = 7;    y = 4 

Substituting in 

x = 20-3y     x = 20-3y 
x = 20-3(7)   x = 20-3(4) 
x = 20-21     x = 20-12 
x = -1        x = 8 

So they intersect at P(-1,7) and Q(8,4) 

 

Now we need to find the equation of another circle that has its
center as the midpoint of PQ, the green circle below:

 

Midpoint of PQ:  %28matrix%281%2C3%2C+++%288%2B%28-1%29%29%2F2%2C%22%2C%22%2C%284%2B7%29%2F2%29%29%22%22=%22%22%28matrix%281%2C3%2C+++7%2F2%2C%22%2C%22%2C11%2F2%29%29 

and has its diameter as the length of PQ and its radius as half that:

 

Therefore its center is %28matrix%281%2C3%2C7%2F2%2C%22%2C%22%2C11%2F2%29%29 and its
radius is one-half the diameter 3sqrt%2810%29%2F2.

Its equation is

%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2

%28x-7%2F2%29%5E2%2B%28y-11%2F2%29%5E2=%283sqrt%2810%29%2F2%29%5E2

%28x-7%2F2%29%5E2%2B%28y-11%2F2%29%5E2=%289%2A10%29%2F4

%28x-7%2F2%29%5E2%2B%28y-11%2F2%29%5E2=90%2F4

%28x-7%2F2%29%5E2%2B%28y-11%2F2%29%5E2=45%2F2

Or you can multiply it out and rearrange to get this:

x%5E2%2By%5E2-7x-11y%2B20+=+0

Edwin

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
your equations are:

x^2 + y^2 - 6x - 8y = 0
x + 3y = 20

solve these 2 equations simultaneously and you will find that you have two point pairs that intersect with the circle and the line.

those point pairs are (-1,7) and (8,4).

find the midpoint of those 2 points and you will find that the midpoint is (7/2, 11/2).

this can also be shown as (3.5, 5.5).

that midpoint is the center of the circle you are looking for.

the diameter of the circle you are looking for is the line segment between (-1,7) and (8,4).

find the length of that line segment and you will find that the length is sqrt(90).

that means that the length of the diameter of your circle with a midpoint at (7/2,11/2) is equal to sqrt(90).

take half of that length and you will find that half the length is equal to sqrt(90)/2.

sqrt(90)/2 is the same as sqrt(90/4).

that's the length of the radius of the circle with the center at (7/2,11/2).

square that radius and you get the equation of the circle with its center at (7/2,11/2).

the equation of that circle will be:

(x-7/2)^2 + (y-11/2)^2 = 90/4.

this can also be shown as (x-3.5)^2 + (y-5.5)^2 = 90/4.

that circle has its diameter as the line segment between the points (-1,7) and (8,4).

the following graph shows the relationships.

$$$

the blue circle is the original circle that was given intersecting with the given line.

the line is black.

the red circle is the circle that has the line as it's diameter.

the center of the blue circle is at (3,4).

that's the center of the original given circle.

the center of the red circle is at (3.5, 5.5).

the blue circle is the original circle.
the red circle is the circle that has the line segment between the points (-1,7) and (8,4) as its diameter.

you find the center of the blue circle by completing the squares on the original equation of the blue circle.

the original equation of the blue circle is:

x^2 + y^2 - 6x - 8y = 0

you associate the x terms and the y terms together to get:

(x^2 - 6x) + (y^2 - 8y) = 0

you complete the squares on the x terms and the y terms to get:

(x-3)^2 + (y-4)^2 = 0 + 9 + 16 which becomes:

(x-3)^2 + (y-4)^2 = 25

that's the standard form of the equation of a circle as (x-h)^2 + (y-k)^2 = r^2

your original circle has a center at (h,k) and a radius of r.

(h,k) is equal to (3,4) and r is equal to sqrt(25) = 5.

when you solved the original two equations of x + 3y = 20 and x^2 + y^2 - 6x - 8y = 0 simultaneously, you found the intersection points of (-1,7) and (8,4).

anyway, without getting into all the gory details, that's how it's done.

if there's any part of this that confuses you, or you don't know how to do, then let me know and i'll take you through it.