Q89 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Mathematics

If $x^2+y^2-a^2+\lambda(x \cos \alpha+y \sin \alpha-p)=0$ is the smallest circle through the points of intersection of $x^2+y^2=a^2$ and $x \cos \alpha+y \sin \alpha=p, 0 < p < a$, then $\lambda=$

  • A
    1
  • B
    $-p$
  • Correct answer
    C
    $-2 p$
  • D
    $-3 p$

Solution

Equation of circle $x^2+y^2-a^2+\lambda$ $(x \cos \alpha+y \sin \alpha-p)=0$ is the smallest circle, then centre $\left(\frac{-\lambda \cos \alpha}{2}, \frac{-\lambda \sin \alpha}{2}\right)$ lies on the line $x \cos \alpha+y \sin \alpha=p$ $\ldots(\mathrm{i})$ $\therefore$ Put centre in the line $(\mathrm{i})$, then we getting $\frac{-\lambda \cos ^2 \alpha}{2}-\frac{\lambda \sin ^2 \alpha}{2}=p \Rightarrow \frac{-\lambda}{2}=p \Rightarrow \lambda=-2 p$

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

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