Consider the family of circles \(x^2+y^2-2 x-2 \lambda y-8=0\) which passes through two fixed points \(A\)…
Consider the family of circles \(x^2+y^2-2 x-2 \lambda y-8=0\) which passes through two fixed points \(A\) and \(B\) distance between them is
- 4
- \(4 \sqrt{2}\)
- 6
- 8
Solution
\(\begin{aligned}
x^2+y^2-2 x-2 \lambda y-8 & =0 \\
\left(x^2+y^2-2 x-8\right)-\lambda(2 y) & =0 \\
x^2+y^2-2 x-8 & =0 \text { and } 2 y=0 \Rightarrow y=0 \\
x^2-2 x-8 & =0
\end{aligned}\)
Distance between points \(A, B=\left|x_1-x_2\right|\)
\(\begin{aligned}
& =\frac{\sqrt{D}}{|a|}=\frac{\sqrt{b^2-4 a c}}{|a|} \\
& =\frac{\sqrt{4-4 \cdot 1(-8)}}{1}=\sqrt{36}=6
\end{aligned}\)
Hence, option (c) is correct.
Asked in: AP EAMCET 2020 (18 Sep Shift 2)
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