+0  
 
0
38
1
avatar

The real numbers $x$ and $y$ satisfy the equation $x^2+y^2=14x+48y-8x-26y+16$. What is the minimum value of $y$?

 Aug 29, 2023
 #1
avatar+4 
+1

First, x^2+y^2=14x+48y-8x-26y+16   rearrange as x^2 -6x + y^2 -22y  = 16 

 

 complete the square on x, y

 

x^2  - 6x + 9  + y^2 -22y + 121  =  16 + 9 + 121

 

(x - 3)^2  + (y -11)^2 = 146

 

This is a  circle centered at (3, 11)  with a radius  of sqrt (146)

 

The minimum value of y =   11 - sqrt (146) ≈   -1.083

 Sep 1, 2023

1 Online Users

avatar