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Zorluk: ZorEquations and Graphs of Circles

A circle in the standard (x,y)(x, y) coordinate plane is defined by the equation x2+y2−6x+4y−12=0x^2 + y^2 - 6x + 4y - 12 = 0. A line LL passes through the center of this circle and is perpendicular to the line with equation 3x−4y=123x - 4y = 12. What is the yy-coordinate of the intersection point of line LL and the circle that has a positive yy-value?

  1. 2Cevap
  2. B
    6
  3. C
    -6
  4. D
    1
  5. E
    5

Cevap

The y-coordinate of the intersection point with a positive y-value is 2.
Completing the square of the circle equation x2+y2−6x+4y−12=0x^2 + y^2 - 6x + 4y - 12 = 0 yields (x−3)2+(y+2)2=25(x-3)^2 + (y+2)^2 = 25, giving center (3,−2)(3, -2) and radius r=5r = 5. The given line 3x−4y=123x - 4y = 12 has a slope of 3/43/4, so a perpendicular line LL has a slope of −4/3-4/3. Since line LL passes through the center (3,−2)(3, -2) and has a slope of −4/3-4/3, moving a distance of 5 units along this line (which corresponds to a horizontal change of ±3\pm 3 and vertical change of ∓4\mp 4) gives the intersection points (6,−6)(6, -6) and (0,2)(0, 2). The y-coordinate of the point with a positive y-value is 2.

Adım Adım Çözüm

1
Complete the square for the circle's equation to find its center and radius.
The equation is rewritten as (x−3)2+(y+2)2=25(x-3)^2 + (y+2)^2 = 25, which represents a circle with center (3,−2)(3, -2) and radius r=5r = 5.
Converting to standard form (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2 gives the circle's center and radius.
2
Find the slope of the given line 3x−4y=123x - 4y = 12 and determine the slope of the perpendicular line LL.
The slope of the given line is 3/43/4. The slope of line LL is the negative reciprocal: −4/3-4/3.
Perpendicular lines have slopes that are negative reciprocals.
3
Find the intersection points of line LL and the circle.
Since line LL passes through the center (3,−2)(3, -2) and has a slope of −4/3-4/3, points on the line at a distance of the radius r=5r=5 are found by moving 3 units horizontally and 4 units vertically. This yields the points (3+3,−2−4)=(6,−6)(3 + 3, -2 - 4) = (6, -6) and (3−3,−2+4)=(0,2)(3 - 3, -2 + 4) = (0, 2).
The intersection points lie exactly one radius away from the center along the line.
4
Identify the y-coordinate with a positive value.
Between the points (6,−6)(6, -6) and (0,2)(0, 2), the point with a positive y-value is (0,2)(0, 2), so the y-coordinate is 2.
The question specifies the intersection point must have a positive y-value.

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Circle standard equations and perpendicular lines on the coordinate plane
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Equations and Graphs of Circles

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