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Zorluk: OrtaEquations and Graphs of Circles

In the standard (x,y)(x, y) coordinate plane, a circle is defined by the equation x2+y2−8x+10y+k=0x^2 + y^2 - 8x + 10y + k = 0, where kk is a constant. If the circle is tangent to the yy-axis, what is the value of kk?

  1. A
    16
  2. 25Cevap
  3. C
    36
  4. D
    37
  5. E
    41

Cevap

The value of kk is 25.
Completing the square for x2+y2−8x+10y+k=0x^2 + y^2 - 8x + 10y + k = 0 gives (x−4)2+(y+5)2=41−k(x - 4)^2 + (y + 5)^2 = 41 - k. The center of the circle is (4,−5)(4, -5). Because the circle is tangent to the yy-axis, its radius is equal to the absolute value of the center's xx-coordinate, which is ∣4∣=4|4| = 4. Therefore, r2=16r^2 = 16. Setting 41−k=1641 - k = 16 gives k=25k = 25.

Adım Adım Çözüm

1
Rewrite the general equation of the circle by completing the square for both xx and yy.
(x2−8x+16)+(y2+10y+25)+k−16−25=0  ⟹  (x−4)2+(y+5)2=41−k(x^2 - 8x + 16) + (y^2 + 10y + 25) + k - 16 - 25 = 0 \implies (x - 4)^2 + (y + 5)^2 = 41 - k
Converting to 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 radius squared r2r^2.
2
Identify the center of the circle and determine the radius using the tangency condition.
Center is (4,−5)(4, -5). Since the circle is tangent to the yy-axis (the line x=0x = 0), the radius rr equals the horizontal distance from the center to the yy-axis: r=∣4∣=4r = |4| = 4.
Tangency to a vertical line means the distance from the center to that line is equal to the radius.
3
Equate the radius squared from step 1 to r2r^2 from step 2 and solve for kk.
41−k=42  ⟹  41−k=16  ⟹  k=2541 - k = 4^2 \implies 41 - k = 16 \implies k = 25
The right side of the standard circle equation represents r2r^2.

Anahtar Kavram

Standard form of a circle equation and conditions for tangency to coordinate axes
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Konular

Equations and Graphs of Circles

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