The number of integral values of lambda for w | vedclass

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(D) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$,where the radius $r = \sqrt{g^2 + f^2 - c}$.Comparing with $x^2 + y^2 + \lambd

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(D) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$,where the radius $r = \sqrt{g^2 + f^2 - c}$.Comparing with $x^2 + y^2 + \lambda x + (1 - \lambda)y + 5 = 0$,we have $g = \frac{\lambda}{2}$,$f = \frac{1 - \lambda}{2}$,and $c = 5$.For the equation to represent a circle,the radius must be real,so $g^2 + f^2 - c > 0$.$\left(\frac{\lambda}{2}\right)^2 + \left(\frac{1 - \lambda}{2}\right)^2 - 5 > 0 \Rightarrow \lambda^2 + (1 - \lambda)^2 - 20 > 0 \Rightarrow 2\lambda^2 - 2\lambda - 19 > 0$.Given the radius $r \leq 5$,we have $r^2 \leq 25$,so $g^2 + f^2 - c \leq 25$.$\frac{\lambda^2}{4} + \frac{(1 - \lambda)^2}{4} - 5 \leq 25 \Rightarrow \lambda^2 + (1 - \lambda)^2 - 20 \leq 100 \Rightarrow 2\lambda^2 - 2\lambda - 119 \leq 0$.Solving $2\lambda^2 - 2\lambda - 119 = 0$ using the quadratic formula: $\lambda = \frac{2 \pm \sqrt{4 - 4(2)(-119)}}{4} = \frac{2 \pm \sqrt{956}}{4} = \frac{1 \pm \sqrt{239}}{2}$.Since $\sqrt{239} \approx 15.46$,the range is $\frac{1 - 15.46}{2} \leq \lambda \leq \frac{1 + 15.46}{2}$,which is $-7.23 \leq \lambda \leq 8.23$.The integral values of $\lambda$ are $\{-7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8\}$.Checking the condition $2\lambda^2 - 2\lambda - 19 > 0$: For $\lambda = 3$,$2(9) - 6 - 19 = -7  0$ (valid). For $\lambda = -2$,$2(4) + 4 - 19 = -7  0$ (valid).The valid integers are $\{-7, -6, -5, -4, -3, 4, 5, 6, 7, 8\}$.There are $5 + 5 = 10$ such values.

The number of integral values of $\lambda$ for which $x^2 + y^2 + \lambda x + (1 - \lambda)y + 5 = 0$ is the equation of a circle whose radius cannot exceed $5$ is:

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