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

Consider the two curves C₁ : y² = 4x, C₂ : x² + y² - 6x + 1 = 0. Then, $C_1: y^2 = 4x \quad (\text{parabola})$ $C_2: x^2 + y^2 - 6x + 1 = 0 \quad (\text{circle})$ $\text{Find: how } C_1 \text{ and } C_2 \text{ meet}$ $C_2: (x-3)^2 + y^2 = 8$ $\text{centre } (3,0), \quad \text{radius } = 2\sqrt{2}$ $y^2 = 4x: \quad x^2 + 4x - 6x + 1 = 0$ $x^2 - 2x + 1 = 0 \Rightarrow (x-1)^2 = 0$ $x = 1 \text{ (double root)}$ $y^2 = 4 \Rightarrow y = \pm 2$ $\text{Points of intersection: } (1, 2), \ (1, -2)$ $\text{Parabola: } y' = \frac{2}{y}; \quad \text{Circle: } y' = \frac{3-x}{y}$ $\text{At } x=1: \ \frac{2}{y} = \frac{3-1}{y} = \frac{2}{y} \ \Rightarrow \text{common tangent}$ $\text{Slopes equal at } (1,\pm2) \Rightarrow \text{tangent at both}$ $\therefore \text{Option } (2): \text{ touch exactly at two points}$