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=
(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 ?
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
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.
To determine the center (h, k) and radius r of the given equation of a circle, you need to rewrite the equation into standard form, which is (x - h)^2 + (y - k)^2 = r^2.
Let's start by completing the square for the x terms:
3x^2 - 12x = 0
3(x^2 - 4x) = 0
To complete the square for the x terms, we need to take half of the coefficient of x (-4) and square it: (-4/2)^2 = 4. We then add and subtract this value inside the parentheses:
3(x^2 - 4x + 4 - 4) = 0
3((x - 2)^2 - 4) = 0
3(x - 2)^2 - 12 = 0
Similarly, complete the square for the y terms:
3y^2 - 30y = 0
3(y^2 - 10y) = 0
Take half of the coefficient of y (-10) and square it: (-10/2)^2 = 25. Add and subtract this value inside the parentheses:
3(y^2 - 10y + 25 - 25) = 0
3((y - 5)^2 - 25) = 0
3(y - 5)^2 - 75 = 0
Now, we can rewrite the entire equation in standard form:
3(x - 2)^2 - 12 + 3(y - 5)^2 - 75 + 84 = 0
3(x - 2)^2 + 3(y - 5)^2 - 3 = 0
(x - 2)^2 + (y - 5)^2 - 1 = 0
Comparing this with the standard form equation, we see that:
h = 2
k = 5
r^2 = 1
To find the radius r, we take the square root of r^2:
r = sqrt(1) = 1
Thus, the center (h, k) of the circle is (2, 5), and the radius r is 1.