The circle x^2 + y^2 - 3x - 4y + 2 = 0 cuts t | vedclass

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(D) Given the equation of the circle is $x^2 + y^2 - 3x - 4y + 2 = 0$.Since the circle cuts the $x$-axis,the $y$-coordinate at these points must be $0

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$(1, 0), (2, 0)$

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(D) Given the equation of the circle is $x^2 + y^2 - 3x - 4y + 2 = 0$.Since the circle cuts the $x$-axis,the $y$-coordinate at these points must be $0$.Substituting $y = 0$ into the equation,we get:$x^2 + 0^2 - 3x - 4(0) + 2 = 0$$x^2 - 3x + 2 = 0$Factoring the quadratic equation:$(x - 1)(x - 2) = 0$Thus,$x = 1$ or $x = 2$.Therefore,the points of intersection are $(1, 0)$ and $(2, 0)$.

The circle $x^2 + y^2 - 3x - 4y + 2 = 0$ cuts the $x$-axis at

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