How do you find center, radius, and intercepts of a circle #3x² + 3y² - 6x + 12y = 0#?

1 Answer
Oct 23, 2016

Answer:

Center: #(1,-2)#
Radius: #sqrt(5)#
Intercepts: #(0,0), (0,-4), (2,0)#

Explanation:

Given
#color(white)("XXX")3x^2+3y^2-6x+12y=0#

Let's start by simplifying by dividing everything by the common factor of #3#
#color(white)("XXX")x^2+y^2-2x+4y=0#

Our target will be to convert this into standard circle form:
#color(white)("XXX")(x-color(red)a)^2+(y-color(blue)b)^2=color(green)r^2#
for a circle with center #(color(red)a,color(blue)b)# and radius #color(green)r#

Regrouping the #x# and #y# terms separately
#color(white)("XXX")(x^2-2xcolor(white)("XX"))+(y^2+4ycolor(white)("XX"))=0#

Completing the square for each
#color(white)("XXX")(x^2-2xcolor(cyan)(+1))+(y^2+4ycolor(purple)(+4))=0color(cyan)(+1)color(purple)(+4)#

Rewriting as squares in standard circle form
#color(white)("XXX")(x-color(red)1)^2+(y-color(blue)(""(-2)))^2=color(green)(""(sqrt(5))^2#

This gives us
#color(white)("XXX")#Center at #(color(red)1,color(blue)(-2))#
#color(white)("XXX")#Radius of #color(green)(sqrt(5))#

To get the intercepts, it is probably easier to work from the earlier equation:
#color(white)("XXX")x^2+y^2-2x+4y=0#

The Y-intercepts occur when #x=0#
#color(white)("XXX")y^2+4y=0#
#color(white)("XXX")y(y+4)=0 #color(white)("XXX")y=0 or y=--4#

The X-intercepts occur when #y=0#
#color(white)("XXX")x^2-2x=0#
#color(white)("XXX")x(x-2)=0#
#color(white)("XXX")x=0 or x=2#

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