Question

3x^2+3y^2-12x-30y+84=0 is the equation of a circle with center
(h,k) and radius r
For:
h=
k=
and r=

Answer 1

(3 x^2 -12 x) + (3 y^2 -30 y) = -84
3( x^2 - 4 x ) + 3 (y^2 -10 y ) =-84
(x^2-4x) + (y^2-10 y) = -28
complete the squares
(x^2-4 x + 4) -4 + (y^2-10 y + 25) -25 = -28
(x-2)^2 + (y-5)^2 = 1 = r^2
got it ?

Answer 2

3 x² + 3 y² -12 x - 30 y + 84 = 0

Subtract 84 to both sides

3 x² + 3 y² -12 x - 30 y = - 84

Divide both sides by 3

x² + y² - 4 x - 10 y = - 28

Add 2² + 5² to both sides

( x² - 4 x + 2² ) + ( y² - 10 y + 5² ) = - 28 + 2² + 5²

( x - 2 )² + ( y - 5² ) = - 28 + 4 + 25

( x - 2 )² + ( y - 5² ) = - 28 + 29

( x - 2 )² + ( y - 5² ) = 1

The standard form equation of circle:

( x - h )² + ( y - k )² = r²

h = 2 , k = 5 , r = 1

Answer 3

My typo:

( x - 2 )² + ( y - 5)² = - 28 + 4 + 25

( x - 2 )² + ( y - 5 )² = - 28 + 29

( x - 2 )² + ( y - 5)² = 1

The same solution as Damon.