Question

15
$x^2 - 6x + y^2 - 4y - 51 = 0$
in the $xy$-plane, the graph of the given equation is a circle. if this circle is inscribed in a square, what is the perimeter of the square?
(a) 16
(b) 32
(c) 64
(d) 204

Explanation:

Step1: Complete the square for x

$x^2 - 6x = (x - 3)^2 - 9$

Step2: Complete the square for y

$y^2 - 4y = (y - 2)^2 - 4$

Step3: Rewrite the circle equation

$(x - 3)^2 - 9 + (y - 2)^2 - 4 - 51 = 0$
$(x - 3)^2 + (y - 2)^2 = 64$
Radius $r = 8$

Step4: Find square side length

Side length = $2r = 16$

Step5: Calculate perimeter

Perimeter = $4×16 = 64$

Answer:

C) 64