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Question 1209537: Let k be a positive real number. The line x + y = k/2 and the circle x^2 + y^2 = 3x - 6y + k are drawn. Find k so that the line is tangent to the circle.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to find the value of k for which the line x + y = k/2 is tangent to the circle x² + y² = 3x - 6y + k:
**1. Rewrite the circle equation:**
Complete the square for both x and y terms in the circle equation:
x² - 3x + y² + 6y = k
(x² - 3x + 9/4) + (y² + 6y + 9) = k + 9/4 + 9
(x - 3/2)² + (y + 3)² = k + 45/4
This shows that the circle has center (3/2, -3) and radius √(k + 45/4).
**2. Express the line in terms of y:**
The line equation x + y = k/2 can be rewritten as y = -x + k/2.
**3. Condition for tangency:**
A line is tangent to a circle if the distance from the center of the circle to the line is equal to the radius of the circle.
**4. Distance from a point to a line:**
The distance from a point (x₁, y₁) to a line ax + by + c = 0 is given by:
Distance = |ax₁ + by₁ + c| / √(a² + b²)
In our case, the point is (3/2, -3) and the line is x + y - k/2 = 0. So, a=1, b=1, c=-k/2, x₁=3/2, and y₁=-3.
Distance = |(1)(3/2) + (1)(-3) - k/2| / √(1² + 1²)
Distance = |3/2 - 3 - k/2| / √2
Distance = |-3/2 - k/2| / √2
Distance = |(-3 - k)/2| / √2
Distance = |k + 3| / (2√2)
**5. Set distance equal to radius:**
Now, set the distance equal to the radius:
|k + 3| / (2√2) = √(k + 45/4)
**6. Solve for k:**
Square both sides to get rid of the square roots:
(k + 3)² / 8 = k + 45/4
(k² + 6k + 9) / 8 = (4k + 45) / 4
k² + 6k + 9 = 2(4k + 45)
k² + 6k + 9 = 8k + 90
k² - 2k - 81 = 0
Use the quadratic formula to solve for k:
k = (-b ± √(b² - 4ac)) / 2a
k = (2 ± √((-2)² - 4(1)(-81))) / 2(1)
k = (2 ± √(4 + 324)) / 2
k = (2 ± √328) / 2
k = (2 ± 2√82) / 2
k = 1 ± √82
Since k must be positive, we take the positive solution:
k = 1 + √82
Therefore, the value of k for which the line is tangent to the circle is 1 + √82.
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