How do you find the center & radius of the circle with the given equation #x²+y²-2x-10y=55# and then graph the equation?

1 Answer
Dec 22, 2015

Answer:

center: #(1,5)#
radius: #9#

Explanation:

The general standard form for the equation of a circle is
#color(white)("XXX")(x-color(orange)(a))^2+(y-color(brown)(b))^2=color(cyan)(r)^2# with center at #(color(orange)(a),color(brown)(b))# and radius #color(cyan)(r)#.

We will attempt to convert the given equation into this form.

Given:
#color(white)("XXX")x^2+y^2-2x-10y=55#
Gather the #x# and #y# terms
#color(white)("XXX")color(blue)(x^2-2x)+color(red)(y^2-10y)=color(green)(55)#
Complete the squares for the #x# and #y# sub-expressions.
#color(white)("XXX")color(blue)(x^2-2x+1)+color(red)(y^2-10y+25)=color(green)(55)+color(blue)(1)+color(red)(25)#
Re-write as squared binomials equal to a squared constant.
#color(white)("XXX")(x-color(orange)(1))^2+(y-color(brown)(5))^2=color(cyan)(9)^2#

Comparing this to the general standard form, we see:
#color(white)("XXX")#the center is at #(color(orange)(1),color(brown)(5))#
and
#color(white)("XXX")#the radius is #color(cyan)(9)#