Question

A circle drawn on a coordinate plane has the equation x2+y2−6x+16y+50.4375=0 . Part A: What is the center of this circle? Express your answer as a coordinate pair.

Answer 1

Answer:

Step-by-step explanation:

First group your x terms and your y terms together, move the constant over to the other side of the equals sign, and then complete the square on each set.

$x^2-6x+y^2+16y=-50.4375$

Completing the square on the x terms:

Take half the linear term, square it and add that squared number to both sides.  Our linear term is 6.  Half of 6 is 3, and 3 squared is 9, so we add in 9 on both sides.

Completing the square on the y terms:

Half of 16 is 8, and 8 squared is 64, so we add in 64 to both sides.

That gives us:

$(x^2-6x+9)+(y^2+16x+64)=-50.4375+9+64$

The purpose of completing the square is to create a perfect square binomial for each set of parenthesis.  It will be in these sets of parenthesis that we find our center.

$(x-3)^2+(y+8)^2=22.5625$

This means that center of the circle is located at (3, -8)