Question

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)
• D. (1, 0), (2, 0)

Answer 1

The correct option is D (1, 0), (2, 0)

Given, equation of circle is $x^2+y^2-3x-4y+2=0$ and it cuts the x-axis.
$\therefore x^2$ + 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).