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assignment 7.2: circles
score: 2/15 answered: 2/15
question 3
$3x^2 + 3y^2 - 12x - 12y + 21 = 0$ is the equation of a circle with center $(h, k)$ and radius $r$ for:
$h = $
and
$k = $
and
$r = $
question help: video 1 video 2 message instructor

Explanation:

Step1: Divide by 3 to simplify

$\frac{3x^2 + 3y^2 - 12x - 12y + 21}{3} = \frac{0}{3}$
$x^2 + y^2 - 4x - 4y + 7 = 0$

Step2: Group x and y terms

$(x^2 - 4x) + (y^2 - 4y) = -7$

Step3: Complete the square for x

Take half of -4: $-2$, square it: $4$. Add to both sides.
$(x^2 - 4x + 4) + (y^2 - 4y) = -7 + 4$
$(x-2)^2 + (y^2 - 4y) = -3$

Step4: Complete the square for y

Take half of -4: $-2$, square it: $4$. Add to both sides.
$(x-2)^2 + (y^2 - 4y + 4) = -3 + 4$
$(x-2)^2 + (y-2)^2 = 1$

Step5: Identify h, k, r

Compare to standard form $(x-h)^2 + (y-k)^2 = r^2$

Answer:

$h = 2$
$k = 2$
$r = 1$