If one of the two circle \(x^2+y^2+\alpha_1(x-y)+c=0\) and \(x^2+y^2+\alpha_2(x-y)+c=0\), lies within the…
If one of the two circle \(x^2+y^2+\alpha_1(x-y)+c=0\) and \(x^2+y^2+\alpha_2(x-y)+c=0\), lies within the other, then (where \(\alpha_1, \alpha_2 \in \mathbf{R}, \alpha_1 \neq \alpha_2\))
\(c < 0\)
\(c=0\)
\(c>0\)
\(c \geq 0\)
Solution
Equation of given circles are
\(\begin{aligned}
& x^2+y^2+\alpha_1(x-y)+C=0 \\
& \text{and } x^2+y^2+\alpha_2(x-y)+C=0
\end{aligned}\)
Since, if one of the two given circles, lies within the other, then
\(C_1 C_2 < \left|r_1-r_2\right|\)
\(\begin{array}{ll}
\text {Where, } & C_1=\left(\frac{-\alpha_1}{2}, \frac{-\alpha_1}{2}\right), C_2=\left(\frac{-\alpha_2}{2}, \frac{-\alpha_2}{2}\right) \\
& r_1=\sqrt{\frac{\alpha_1^2}{2}-C} \text { and } r_2=\sqrt{\frac{\alpha_2^2}{2}-C} \\
\therefore & \sqrt{\frac{\left(\alpha_1-\alpha_2\right)^2}{2}} < \left|\sqrt{\frac{\alpha_1^2}{2}-C}-\sqrt{\frac{\alpha_2^2}{2}-C}\right|
\end{array}\)
\(\Rightarrow \frac{\alpha_1^2+\alpha_2^2-2 \alpha_1 a_2}{2} < \left(\frac{\alpha_1^2}{2}-C\right) +\left(\frac{\alpha_2^2}{2}-C\right) -2 \sqrt{\frac{\alpha_1^2}{2}-C} \sqrt{\frac{\alpha_2^2}{2}-C}\)
\(\Rightarrow \quad-\alpha_1 \alpha_2 < -2 C-\sqrt{\alpha_1^2-2 C} \sqrt{\alpha_2^2-2 C}\)
\(\Rightarrow \quad\left(\alpha_1^2-2 C\right)\left(\alpha_2^2-2 C\right) < \left(\alpha_1 \alpha_2-2 C\right)^2\)
\(\Rightarrow \alpha_1^2 \alpha_2^2-2\left(\alpha_1^2+\alpha_2^2\right) C+4 C^2 < \alpha_1^2 \alpha_2^2-4 C \alpha_1 \alpha_2+4 C^2\)
\(\Rightarrow 2 C\left[\alpha_1^2+\alpha_2^2-2 \alpha_1 \alpha_2\right] > 0\)
\(\Rightarrow C\left(\alpha_1-\alpha_2\right)^2 > 0 \Rightarrow C > 0\) as \(\alpha_1 \neq \alpha_2\)
Hence, option (c) is correct.