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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 x?

 Jul 28, 2019
 #1
avatar+106539 
+1

x^2 + y^2  = 14x + 48y         rearrange as

 

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

 

x^2 -14x + 49 + y^2 - 48y + 576  =  49 + 576        factor  and simplify

 

(x -7)^2  + ( y - 24)^2  =  625

 

This is a circle centered at  ( 7 , 24)   with  a radius of √625  =  25

 

The minimum value  for x  will be :     x coordinate of the center  minus the radius  =

 

7 - 25   =    -18

 

 

cool cool cool

 Jul 28, 2019
 #2
avatar+1195 
+1

Thanks CPhill.

 Jul 28, 2019

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