We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+1
283
2
avatar+814 

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?

 Aug 25, 2018
 #1
avatar+100587 
+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

 Aug 25, 2018
 #2
avatar+814 
+2

Correct! Thanks!

mathtoo  Aug 25, 2018

8 Online Users

avatar