+0  
 
0
60
1
avatar

Write the general equation for the circle that passes through the points (1,1) (1,3) (9,2)

 Mar 19, 2020
 #1
avatar+111360 
+1

Let ( h,k) be  the center  of the circle

 

Since  the square of the radius  =  the square of the radius we can set up this equation

 

(h - 1)^2  + ( k - 1)^2  =  (h - 1)^2  + (k - 3)^2     simplify

 

(k -1)^2  = (k - 3)^2 

 

k^2  - 2k + 1  =  k^2  - 6k + 9

 

-2k + 1 =  -6k + 9

 

4k  =  8

 

k  = 2

 

Then we have that

 

(h -1)^2 + (k-1)^2  =  ( h - 9)^2  + (k - 2)^2

 

Subbing in for k we have

 

(h -1)^2  + (2 -1)^2  = (h -9)^2  + (2-2)^2    simplify

 

h^2 - 2h + 1  + 1  =  h^2 - 18h + 81

 

-2h + 2 =  -18h+ 81

 

16h  =  79

 

h = 79/16

 

So  the center is  (79/16 ,2)

 

And  the  radius can be  found as

 

(79/16 - 1)^2  + (2 -1)^2  =  r^2

 

(63/16)^2  + 1  = r^2

 

4225/256  =  r^2

 

So the equation is

 

( x - 79/16)^2 + (y - 2)^2  = 4225/256

 

 

cool cool cool

 Mar 19, 2020

16 Online Users

avatar