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Zorluk: OrtaEquations and Graphs of Circles

A circle in the standard (x,y)(x, y) coordinate plane is described by the equation x2+y2−6x+10y+18=0x^2 + y^2 - 6x + 10y + 18 = 0. A second circle is concentric with the first circle (meaning they share the exact same center) but has an area that is 44 times the area of the first circle. Which of the following is an equation of the second circle?

  1. (x−3)2+(y+5)2=64(x - 3)^2 + (y + 5)^2 = 64Cevap
  2. B
    (x+3)2+(y−5)2=64(x + 3)^2 + (y - 5)^2 = 64
  3. C
    (x−3)2+(y+5)2=32(x - 3)^2 + (y + 5)^2 = 32
  4. D
    (x+3)2+(y−5)2=32(x + 3)^2 + (y - 5)^2 = 32
  5. E
    (x−3)2+(y+5)2=16(x - 3)^2 + (y + 5)^2 = 16

Cevap

The equation of the second circle is (x−3)2+(y+5)2=64(x - 3)^2 + (y + 5)^2 = 64.
Completing the square on x2+y2−6x+10y+18=0x^2 + y^2 - 6x + 10y + 18 = 0 yields (x−3)2+(y+5)2=16(x - 3)^2 + (y + 5)^2 = 16, identifying the center as (3,−5)(3, -5) and r12=16r_1^2 = 16. Concentric circles share the center (3,−5)(3, -5). Quadrupling the area means the new area is 4×16π=64π4 \times 16\pi = 64\pi, so r22=64r_2^2 = 64. Plugging the center and new r2r^2 into the standard equation gives (x−3)2+(y+5)2=64(x - 3)^2 + (y + 5)^2 = 64.

Adım Adım Çözüm

1
Convert the given circle equation to standard form by completing the square for both xx and yy.
Rearranging terms: (x2−6x)+(y2+10y)=−18(x^2 - 6x) + (y^2 + 10y) = -18.
Adding (−6/2)2=9( -6/2 )^2 = 9 and (10/2)2=25( 10/2 )^2 = 25 to both sides yields:
(x−3)2+(y+5)2=−18+9+25=16(x - 3)^2 + (y + 5)^2 = -18 + 9 + 25 = 16.
Standard form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2 reveals the center (h,k)(h, k) and squared radius r2r^2.
2
Identify the center and radius of the original circle.
Center = (3,−5)(3, -5) and radius squared r12=16r_1^2 = 16 (so r1=4r_1 = 4).
Concentric circles share the exact same center (h,k)=(3,−5)(h, k) = (3, -5).
3
Determine the radius squared of the second circle based on the area constraint.
The area of a circle is A=πr2A = \pi r^2. Since A2=4A1A_2 = 4 A_1, we have πr22=4(πr12)  ⟹  r22=4r12=4(16)=64\pi r_2^2 = 4 (\pi r_1^2) \implies r_2^2 = 4 r_1^2 = 4(16) = 64.
Quadrupling the area quadruples the value of r2r^2.
4
Write the standard form equation for the second circle using center (3,−5)(3, -5) and r22=64r_2^2 = 64.
(x−3)2+(y+5)2=64(x - 3)^2 + (y + 5)^2 = 64.
Substituting h=3h = 3, k=−5k = -5, and r22=64r_2^2 = 64 into (x−h)2+(y−k)2=r22(x - h)^2 + (y - k)^2 = r_2^2.

Anahtar Kavram

Standard form of a circle equation (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2 and area relationship A=πr2A = \pi r^2
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Konular

Equations and Graphs of Circles

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