Consider the two curves $C_{1} : y^{2} = 4 x, C_{2} : x^{2} + y^{2} - 6 x + 1 = 0$ . Then which of the following option(s) is/are correct?

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$C_{1}$ and $C_{2}$ touch each other exactly at two points.

$C_{1}$ and $C_{2}$ touch each other exactly at one points.

Equation of the tangent at $(1, 2)$ to $C_{1}$ and $C_{2}$ both is $x - 2 y + 1 = 0$ and at $(1, - 2)$ is $2 x + y + 1 = 0$ .

Equation of the tangent at $(1, 2)$ to $C_{1}$ and $C_{2}$ both is $x - y + 1 = 0$ and at $(1, - 2)$ is $x + y + 1 = 0$ .

answer is A, B.

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Detailed Solution

Solving for the points of intersection, we have $x^{2} + 4 x - 6 x + 1 = 0$

$\Rightarrow {(x - 1)}^{2} = 0 \Rightarrow x = 1 \Rightarrow y = \pm 2$

Thus, the two curves meet at (1,2) and (1,-2).
Tangent at (1,2) to $C_{1} : y^{2} = 4 x i s y (2) = 2 (x + 1)$
$\Rightarrow x - y + 1 = 0$
Tangent at (1,2) to $C_{2} : x^{2} + y^{2} - 6 x + 1 = 0$ is $x + 2 y - 3 (x + 1) + 1 = 0$
or $x - y + 1 = 0$ same as the tangent to the curve $C_{1}$ .
Similarly, the tangent at the point (1,-2) to the two curves is $x + y + 1 = 0$
Statement II is true and hence, Statement I is also true.