Q58 / 160 · AP EAMCET 2020 (18 Sep Shift 2) · Mathematics

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

  • A
    4
  • B
    \(4 \sqrt{2}\)
  • Correct answer
    C
    6
  • D
    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.

Hence, the correct answer is option: 'C'.

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