Question
The number of integral values of $\lambda$ for which the equation $\text{x}^2+\text{y}^2+\lambda+(1-\lambda)\text{y}+5=0$ is the equation of a circle whose radius cannot exceed 5, is:
- 14
- 18
- 16
- None of these
Solution:
$\sqrt{\Big(\frac{-\lambda}{2}^2\Big)+\Big(\frac{\lambda-1}{2}\Big)^2-5}\leq5$
$\Rightarrow\Big(\frac{-\lambda}{2}^2\Big)+\Big(\frac{\lambda-1}{2}\Big)\leq30$
$\lambda^2+(\lambda-1)^2\leq120$
$\Rightarrow2\lambda^2-2\lambda-199\leq0$
Using quadratic formula:
$\Rightarrow\lambda=\frac{2\pm\sqrt{2^2-4(2)(-119)}}{2(2)}$
$\Rightarrow\lambda=\frac{2\pm\sqrt{956}}{4}$
$\Rightarrow\lambda=\frac{1\pm\sqrt{239}}{2}$
$\Rightarrow\lambda=-7.23,\ 8.23$
$\Rightarrow-7.23\leq\lambda\leq8.23$
$\Rightarrow\lambda=-7,\ -6,\ -5,\ -4,\ -3,\ -2,\ -1\\0,\ 1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8,\ $ $(\text{if}\ \lambda\in\text{Z})$
Thus, the number of integral values of $\lambda$ is 16.
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If slope of a line is 4 and y-intercept made by the line is 2 then the equation of line will be:
-2
0
$\frac{1}{2}$
does not exist