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Question: The number of integral value of l for which x<sup>2</sup> + y<sup>2</sup> + 2l x + 2(1 –l)y –1 = 0 ...

The number of integral value of l for which

x2 + y2 + 2l x + 2(1 –l)y –1 = 0 is the equation of a circle whose radius can not exceed 3

A

3

B

4

C

5

D

None

Answer

4

Explanation

Solution

λ2+(1λ)2+1\sqrt { \lambda ^ { 2 } + ( 1 - \lambda ) ^ { 2 } + 1 }< 3

l2 + l2 –2l + 1 + 1 < 9

2l2 –2l – 7 < 0

1+152\frac { 1 + \sqrt { 15 } } { 2 }

–1.45 < l < 2. 45

–1, 0, 1, 2