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Zorluk: OrtaEquations of Circles in the Coordinate Plane

In the xyxy-plane, a circle is defined by the equation x2+y2+kx−8y=9x^2 + y^2 + kx - 8y = 9, where kk is a constant. If the center of the circle lies on the line y=2xy = 2x, what is the value of kk?

  1. A
    -16
  2. -4Cevap
  3. C
    -2
  4. D
    4

Cevap

-4
The correct answer is the value that satisfies the relation between the circle's center coordinates and the given line. By completing the square on the circle's equation, we rewrite it as (x+k/2)2+(y−4)2=25+k2/4(x + k/2)^2 + (y - 4)^2 = 25 + k^2/4. This reveals the center of the circle is at the point (-k/2, 4). Substituting these coordinates into the equation of the line, which is y = 2x, gives the equation 4 = 2(-k/2). Simplifying this relation gives 4 = -k, which yields k = -4.

Adım Adım Çözüm

1
Group the x and y terms of the circle equation.
(x2+kx)+(y2−8y)=9(x^2 + kx) + (y^2 - 8y) = 9
Grouping terms helps in completing the square for x and y separately.
2
Complete the square for both the x and y quadratic expressions.
(x+k2)2−k24+(y−4)2−16=9(x + \frac{k}{2})^2 - \frac{k^2}{4} + (y - 4)^2 - 16 = 9, which simplifies to (x+k2)2+(y−4)2=25+k24(x + \frac{k}{2})^2 + (y - 4)^2 = 25 + \frac{k^2}{4}
This puts the equation into the standard circle form (x−h)2+(y−j)2=r2(x - h)^2 + (y - j)^2 = r^2.
3
Identify the coordinates of the center of the circle.
Center: (h,j)=(−k2,4)(h, j) = (-\frac{k}{2}, 4)
In standard form, the terms (x−h)2(x - h)^2 and (y−j)2(y - j)^2 correspond to a center at (h,j)(h, j).
4
Substitute the center coordinates into the line equation y=2xy = 2x to solve for kk.
4=2(−k2)  ⟹  4=−k  ⟹  k=−44 = 2(-\frac{k}{2}) \implies 4 = -k \implies k = -4
Since the center lies on the line, the coordinates of the center must satisfy the line's equation.

Anahtar Kavram

Completing the square to find the center of a circle in the coordinate plane and using coordinate geometry relations.

Alternatif Yöntem

Instead of completing the square fully, recall that for any circle equation of the form x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0, the center coordinates (h,j)(h, j) are given by (−D/2,−E/2)(-D/2, -E/2). In the given equation, D=kD = k and E=−8E = -8. Therefore, the center is (−k2,4)(-\frac{k}{2}, 4). Substituting these coordinates directly into the line equation y=2xy = 2x yields 4=2(−k2)4 = 2(-\frac{k}{2}), which simplifies to k=−4k = -4.
Tahmini Süre:1m 30s
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Konular

Equations of Circles in the Coordinate Plane

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