Math  /  Geometry

Questionx26x+y24y51=0x 2-6 x+y 2-4 y-51=0
In the xyx y-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

Studdy Solution

STEP 1

1. The given equation represents a circle in the xyxy-plane.
2. The circle is inscribed in a square.
3. We need to find the perimeter of the square.

STEP 2

1. Rewrite the given equation in standard form of a circle.
2. Determine the radius of the circle.
3. Calculate the side length of the square.
4. Calculate the perimeter of the square.

STEP 3

Rewrite the given equation in standard form of a circle by completing the square.
The given equation is:
x26x+y24y51=0 x^2 - 6x + y^2 - 4y - 51 = 0
Complete the square for the xx terms:
x26x x^2 - 6x
Add and subtract (62)2=9\left(\frac{6}{2}\right)^2 = 9:
x26x+99 x^2 - 6x + 9 - 9
Complete the square for the yy terms:
y24y y^2 - 4y
Add and subtract (42)2=4\left(\frac{4}{2}\right)^2 = 4:
y24y+44 y^2 - 4y + 4 - 4
Rewrite the equation:
(x3)29+(y2)2451=0 (x - 3)^2 - 9 + (y - 2)^2 - 4 - 51 = 0
(x3)2+(y2)2=64 (x - 3)^2 + (y - 2)^2 = 64

STEP 4

Determine the radius of the circle.
The standard form of a circle is:
(xh)2+(yk)2=r2 (x - h)^2 + (y - k)^2 = r^2
From the equation (x3)2+(y2)2=64(x - 3)^2 + (y - 2)^2 = 64, we have:
r2=64 r^2 = 64
r=64=8 r = \sqrt{64} = 8

STEP 5

Calculate the side length of the square.
Since the circle is inscribed in the square, the diameter of the circle is equal to the side length of the square:
Diameter=2×Radius=2×8=16 \text{Diameter} = 2 \times \text{Radius} = 2 \times 8 = 16

STEP 6

Calculate the perimeter of the square.
The perimeter of a square is given by:
Perimeter=4×Side Length \text{Perimeter} = 4 \times \text{Side Length}
Perimeter=4×16=64 \text{Perimeter} = 4 \times 16 = 64
The perimeter of the square is:
64 \boxed{64}

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