How do you graph #(x-2)^2 + (y+5)^2#?

1 Answer
Oct 28, 2015


Transposed circle.


This is the equation of a circle so you should really have stated it as: # (x-2)^2 +(y+5)^2 = r^2#

This equation is derived from Pythagoras relating the sides of a triangle where #("hypotenuse")^2 = ("opposite")^2 + ("adjacent")^2 #

Changing the values of x and y by constant values means that you have moved the centre of the circle away from the origin. The origin being #( x , y ) -> (0,0)#

So the centre has moved to -2 on the x-axis and to +5 on the y axis.

To actually draw the graph you need to know the value of the radius #r#. If #r# is not given then make one up that is easy to use.

Circle at the origin. Move every point to #(x-2, y+5)#
So put the needle of your compass at #(-2,5)# and draw your circle.

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