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avatar+111 

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+103689 
+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+103122 
+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

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