How do you solve #2( x + 2) ^ { 2} + 2( y - 3) ^ { 2} = 10#?

1 Answer
Sep 27, 2017

It has an infinite number of solutions; it is a circle with its center located at #(-2,3)# and a #"radius "=sqrt5#. Every point on the circle is a solution.

Explanation:

Given: #2( x + 2)^2 + 2( y - 3)^2 = 10#

Divide both sides of the equation by 2:

#( x + 2)^2 + ( y - 3)^2 = 5#

The standard form is #(x - h)^2 + (y-k)^2 = r^2#, therefore, we write the +2 as a negative number:

#( x - (-2))^2 + ( y - 3)^2 = 5#

And we write the 5 as the square of a square root:

#( x - (-2))^2 + ( y - 3)^2 = (sqrt(5))^2#

This allows us to easily observe that the center is #(-2,3)# and the radius is #sqrt5#