How do you find the center and radius for #x^2 + y^2 - 6x - 4y - 12 = 0#?

2 Answers
Nov 1, 2016

Answer:

Use the quadratic 'Complete the Square' method

Explanation:

#x^2 - 6x +y^2 - 4y = 12

Then take 1/2 of the 'b' term for both quadratic expressions, square those values and add them to both sides.

#x^2 -6x + 9 + y^2 - 4y + 4 = 12 + 9 + 4

(x - 3)^2 + (y -2)^2 = 25

Circle centered at (3,2) with radius = 5

Nov 1, 2016

Answer:

Center: #(3,2)color(white)("XXXXX")#Radius: #5#
#color(white)("XXX")#(see below for method)

Explanation:

An equation in the form:
#color(white)("XXX")(x-color(red)a)^2+(y-color(blue)b)^2=color(magenta)r^2#
is the standard form for the equation of a circle with center #(color(red)a,color(blue)b)# and radius #color(magenta)r#.

Lets try to convert the given equation:
#color(white)("XXX")x^2+y^2-6x-4y-12=0#
into the standard form for the equation of a circle.

Group the #x# and #y# terms separately and "move" the constant to the right side of the equation:
#color(white)("XXX")x^2-6xcolor(white)(+3^2)+y^2-4ycolor(white)(+2^2)=12#

Complete the square for each of #x# and #y#
#color(white)("XXX")x^2-6xcolor(magenta)(+3^2)+y^2-4ycolor(orange)(+2^2)=12color(magenta)(+3^2)color(orange)(+2^2)#

Write the left side as the sum of two squared binomials
and simplify the result on the right side
#color(white)("XXX")(x-3)^2+(y-2)^2=25#

Express the right side as a square.
#color(white)("XXX")(x-color(red)3)^2+(y-color(blue)2)^2=color(green)5^2#
...the equation for a circle with center #(color(red)3,color(blue)2)# and radius #color(green)5#