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  
 
0
205
2
avatar+124 

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

 Feb 8, 2019
 #1
avatar+105534 
+2

This is going to be a circe, just did a similar one for you.

Once you have the centre of the circle the minimim x value will be the centre x value minus the radius.

 Feb 8, 2019
 #2
avatar+104723 
+3

x^2 -14x + y^2 - 48y = 0

 

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

 

(x - 7)^2 + ( y - 24)^2 =  625

 

x is minimized/maximized when y = 24

 

(x - 7)^2 =  625        take both roots

 

x - 7  =  25          or        x  - 7 =  -25

x = 32                 or        x =  - 18 

 

So....x = -18  is the minimum value for x

 

 

cool cool cool

 Feb 8, 2019

41 Online Users

avatar
avatar
avatar