MCQ
The circle ${x^2} + {y^2} - 3x - 4y + 2 = 0$ cuts $x$-axis at
- A$(2,\,0),( - 3,\,0)$
- B$(3,\,0),(4,\,0)$
- C$(1,\,0),( - 1,\,0)$
- ✓$(1,\,0),(2,\,0)$
$\therefore $${x^2} + 0 - 3x + 2 = 0$ or ${x^2} - 3x + 2 = 0$ or $(x - 1)\,(x - 2) = 0$ or $x = 1,\,2.$
Therefore the points are $(1,0)$ and $(2, 0)$.
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