You can put this solution on YOUR website! A circle is defined by the following equation :
3x^2+3y^2-24x+42y+183=0
What is the radius?
The standard form of the equation of a circle is , where the center is (h,k) and the radius is r.
First divide the whole equation by 3 because x^2 and y^2 have to have a coefficient of 1 in order to be in standard form.
3x^2/3+3y^2/3-24x/3+42y/3+183/3=0/3
x^2+y^2-8x+14y+61=0
Group the x terms and y terms and subtract 61 from both sides of the equation
x^2-8x+__+y^2+14y+___=-61+___+___
We need to complete the square by taking half of the coefficients of x and y and squaring them and adding them to both sides of the equation.
x^2-8x+(-8/2)^2+y^2+14y+(14/2)^2=-61+(-8/2)^2+(14/2)^2
x^2-8x+(-4)^2+y^2+14y+(7)^2=-61+(-4)^2+(7)^2
(x-4)^2+(y+7)^2=-61+16+49
(x-4)^2+(y+7)^2=4
(x-4)^2+(y+7)^2=2^2
The radius is 2
You didn't ask for the center, but it's (4,-7)
Happy Calculating!!!
(