Circle \( \Large x^{2}+y^{2}-2x-nx-1=0 \) passes through two fixed points, co-ordinates of the points are


A) \( \Large \left(0,\ \pm 1\right) \)

B) \( \Large \left(\pm\ 1,\ 0\right) \)

C) \( \Large \left(0,\ 1\right)\ and\ \left(0,\ 2\right) \)

D) \( \Large \left(0,\ -1\right)\ and\ \left(0,\ -2\right) \)

Correct Answer:
A) \( \Large \left(0,\ \pm 1\right) \)

Description for Correct answer:
\( \Large \left(x^{2}+y^{2}-2x-1\right)+nx=0 \), they pass through intersection points of line x = 0 and circle.

\( \Large x^{2}+y^{2}-2x-1=0 \)

=> \( \Large y = \pm 1 \)

Therefore, Required points are \( \Large \left(0,\ \pm 1\right) \)

Part of solved Circles questions and answers : >> Elementary Mathematics >> Circles








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