Answer to Question #291827 in Analytic Geometry for Bonet

Question

How far from the y-axis is the center of the curve 2x²+2y²+10x-6y-55=0

Answer 1

Let us find how far from the $y$-axis is the center of the curve $2x^2+2y^2+10x-6y-55=0.$

This equation is equivalent to $x^2+y^2+5x-3y=27.5,$ and hence to $x^2+5x+(2.5)^2+y^2-3y+(1.5)^2=27.5+(2.5)^2+(1.5)^2.$

The last equation is equivalent to $(x+2.5)^2+(y-1.5)^2=36.$

Therefore, the point $(-2.5,1.5)$ is the center of the circle $2x^2+2y^2+10x-6y-55=0.$

It follows that that the distance between the y-axis and the center $(-2.5,1.5)$ is equal to $2.5.$