Consider the two curves C_1: y^2=4x and C_2: | vedclass

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(B) To find the points of intersection,substitute $y^2 = 4x$ into the equation of the circle $x^2 + y^2 - 6x + 1 = 0$.This gives $x^2 + 4x - 6x + 1 =

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$C_1$ and $C_2$ touch each other exactly at two points.

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(B) To find the points of intersection,substitute $y^2 = 4x$ into the equation of the circle $x^2 + y^2 - 6x + 1 = 0$.This gives $x^2 + 4x - 6x + 1 = 0$,which simplifies to $x^2 - 2x + 1 = 0$.This is $(x - 1)^2 = 0$,so $x = 1$.Substituting $x = 1$ into $y^2 = 4x$,we get $y^2 = 4$,which means $y = 2$ or $y = -2$.Thus,the curves intersect at the points $(1, 2)$ and $(1, -2)$.To check if they touch,we compare the slopes of the tangents at these points.For the parabola $y^2 = 4x$,differentiating gives $2y \frac{dy}{dx} = 4$,so $\frac{dy}{dx} = \frac{2}{y}$. At $(1, 2)$,slope $m_1 = 1$. At $(1, -2)$,slope $m_1 = -1$.For the circle $x^2 + y^2 - 6x + 1 = 0$,differentiating gives $2x + 2y \frac{dy}{dx} - 6 = 0$,so $\frac{dy}{dx} = \frac{3 - x}{y}$. At $(1, 2)$,slope $m_2 = \frac{3 - 1}{2} = 1$. At $(1, -2)$,slope $m_2 = \frac{3 - 1}{-2} = -1$.Since the slopes are equal at both points,the curves touch each other at exactly two points.

Consider the two curves $C_1: y^2=4x$ and $C_2: x^2+y^2-6x+1=0$. Then,

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