+0  
 
+1
72
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avatar+621 

Let (x,y) be an ordered pair of real numbers that satisfies the equation \(x^2+y^2=14x+48y\). What is the minimum value of y?

mathtoo  Aug 25, 2018
 #1
avatar+91027 
+3

x^2 + y^2  = 14x + 48y     rearrange

 

x^2 - 14x + y^2 - 48y  = 0    complete the square on x, y

 

(x^2  - 14x + 49) + (y^2 - 48y + 576)  =  49 + 576

 

(x - 7)^2  +  ( y - 24)^2  =  625

 

(x - 7)^2  + (y - 24)^2  = 25^2

 

This is a circle  with a center at  ( 7, 24)  and a radius of 25

 

The minimum y value will occur  at  ( 7, 24 - 25)  = ( 7, -1)

 

So...the  minimum y value  =  -1

 

 

cool cool cool

CPhill  Aug 25, 2018
 #2
avatar+621 
+2

Correct! Thanks!

mathtoo  Aug 25, 2018

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