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If the circles x2 + y2 = 2 and x2 + y2 - 4x - 4y + λ = 0 have exactly three real common tangents then λ =
- a)-10
- b)6
- c)-6
- d)10
Correct answer is option 'B'. Can you explain this answer?
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If the circles x2 + y2 = 2 and x2 + y2 - 4x - 4y + λ = 0 have e...
C1C2 = r1+ r2
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If the circles x2 + y2 = 2 and x2 + y2 - 4x - 4y + λ = 0 have e...
Given Information:
- The two circles are x^2 + y^2 = 2 and x^2 + y^2 - 4x - 4y + λ = 0.
- The circles have exactly three real common tangents.
Solution:
Step 1: Finding the Center and Radius of the Circles
- The first circle has the center at (0,0) and radius √2.
- The second circle can be rewritten as (x-2)^2 + (y-2)^2 = 2 + 2^2 - λ = 6 - λ. This circle has the center at (2,2) and radius √(6 - λ).
Step 2: Relationship between Radii for Common Tangents
- For three real common tangents, the circles must be externally tangent to each other.
- The sum of the radii of the circles must be equal to the distance between their centers.
- Therefore, √2 + √(6 - λ) = √((2-0)^2 + (2-0)^2) = √8 = 2√2.
Step 3: Solving for λ
- √2 + √(6 - λ) = 2√2
- √(6 - λ) = √2
- 6 - λ = 2
- λ = 6 - 2
- λ = 4.
Conclusion:
- The value of λ that satisfies the condition of having exactly three real common tangents is 4.
- Therefore, the correct answer is option B: 6.
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If the circles x2 + y2 = 2 and x2 + y2 - 4x - 4y + λ = 0 have exactly three real common tangents then λ =a)-10b)6c)-6d)10Correct answer is option 'B'. Can you explain this answer?
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