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The number of integral values of λ for which x2 + y2 + λx + (1 – λ)y + 5 = 0 is the equation of a circle whose radius cannot exceed 5, is
- a)14
- b)18
- c)16
- d)None
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
The number of integral values of λ for which x2 + y2 + λ...
Radius ≤ 5
∴ λ = -7, - 6, - 5, .......,7, 8 , in all 16 values
∴ λ = -7, - 6, - 5, .......,7, 8 , in all 16 values
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The number of integral values of λ for which x2 + y2 + λ...
Let's call the number of integral values of $x$ and $y$ that satisfy the equation $x^2 - 100y^2 = 1$ as $N$.
First, let's consider the equation $x^2 - 1 = 100y^2$. This equation is equivalent to $(x-1)(x+1) = 100y^2$. Since $x-1$ and $x+1$ are consecutive integers, one of them must be divisible by $4$ and the other must be divisible by $2$. Therefore, we have two cases to consider:
Case 1: $x-1$ is divisible by $4$ and $x+1$ is divisible by $2$. Let $x-1 = 4a$ and $x+1 = 2b$, where $a$ and $b$ are integers. Substituting these into the equation, we have $(4a)(2b) = 100y^2$, which simplifies to $ab = 25y^2$. The only possible values of $a$ and $b$ that satisfy this equation are $(1, 25)$ and $(25, 1)$. Therefore, we have two solutions for this case: $(x,y) = (5, \pm 1)$.
Case 2: $x-1$ is divisible by $2$ and $x+1$ is divisible by $4$. Let $x-1 = 2a$ and $x+1 = 4b$, where $a$ and $b$ are integers. Substituting these into the equation, we have $(2a)(4b) = 100y^2$, which simplifies to $ab = 50y^2$. The possible values of $a$ and $b$ that satisfy this equation are $(1, 50)$, $(2, 25)$, $(5, 10)$, $(10, 5)$, $(25, 2)$, and $(50, 1)$. Therefore, we have six solutions for this case: $(x,y) = (3, \pm 1)$, $(7, \pm 3)$, and $(13, \pm 5)$.
In total, we have $2 + 6 = \boxed{8}$ integral solutions for $x$ and $y$.
First, let's consider the equation $x^2 - 1 = 100y^2$. This equation is equivalent to $(x-1)(x+1) = 100y^2$. Since $x-1$ and $x+1$ are consecutive integers, one of them must be divisible by $4$ and the other must be divisible by $2$. Therefore, we have two cases to consider:
Case 1: $x-1$ is divisible by $4$ and $x+1$ is divisible by $2$. Let $x-1 = 4a$ and $x+1 = 2b$, where $a$ and $b$ are integers. Substituting these into the equation, we have $(4a)(2b) = 100y^2$, which simplifies to $ab = 25y^2$. The only possible values of $a$ and $b$ that satisfy this equation are $(1, 25)$ and $(25, 1)$. Therefore, we have two solutions for this case: $(x,y) = (5, \pm 1)$.
Case 2: $x-1$ is divisible by $2$ and $x+1$ is divisible by $4$. Let $x-1 = 2a$ and $x+1 = 4b$, where $a$ and $b$ are integers. Substituting these into the equation, we have $(2a)(4b) = 100y^2$, which simplifies to $ab = 50y^2$. The possible values of $a$ and $b$ that satisfy this equation are $(1, 50)$, $(2, 25)$, $(5, 10)$, $(10, 5)$, $(25, 2)$, and $(50, 1)$. Therefore, we have six solutions for this case: $(x,y) = (3, \pm 1)$, $(7, \pm 3)$, and $(13, \pm 5)$.
In total, we have $2 + 6 = \boxed{8}$ integral solutions for $x$ and $y$.
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The number of integral values of λ for which x2 + y2 + λx + (1 – λ)y + 5 = 0 is the equation of a circle whose radius cannot exceed 5, isa)14b)18c)16d)NoneCorrect answer is option 'C'. Can you explain this answer?
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