This problem requires us to find common tangents to a parabola and a circle. The general strategy involves expressing the equation of a tangent to one curve in terms of its slope, and then applying the tangency condition for the second curve. This will lead to an equation whose solutions are the slopes of the common tangents.

Key Concepts and Formulas

1. Equation of a Tangent to a Parabola in Slope Form: For a parabola of the form $y^2 = 4ax$, the equation of a tangent with slope $m$ is given by: $y = mx + \frac{a}{m}$ This form is particularly useful when the slope $m$ is the unknown we are trying to determine.

2. Condition for Tangency to a Circle: A line $Ax + By + C = 0$ is tangent to a circle with center $(x_0, y_0)$ and radius $R$ if and only if the perpendicular distance from the center of the circle to the line is equal to the radius. The formula for the perpendicular distance $d$ from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is: $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$ For tangency, we set $d = R$.

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Step 1: Determine the Equation of a Tangent to the Parabola

• Given Parabola: The equation of the parabola is $y^2 = x$.
• Identify 'a': We compare this to the standard form of a parabola, $y^2 = 4ax$. By comparing $y^2 = x$ with $y^2 = 4ax$, we find that $4a = 1$, which implies $a = \frac{1}{4}$.
• Formulate Tangent Equation: Now, we substitute this value of $a$ into the slope form of the tangent equation for a parabola, $y = mx + \frac{a}{m}$: $y = mx + \frac{1/4}{m}$ $y = mx + \frac{1}{4m} \quad \text{(Equation 1)}$ Why this step? We are looking for lines that are tangent to both the parabola and the circle. By starting with the general tangent equation for the parabola, we parameterize all possible tangents to the parabola in terms of their slope $m$. Our next step will be to filter these lines to find those that also satisfy the tangency condition for the circle.

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Step 2: Apply the Tangency Condition for the Circle

• Given Circle: The equation of the circle is $x^2 + y^2 = 2$.
• Identify Center and Radius: We compare this to the standard form of a circle centered at the origin, $x^2 + y^2 = R^2$. The center of the circle is $(x_0, y_0) = (0, 0)$. The radius of the circle is $R = \sqrt{2}$.
• Rewrite Tangent Equation in General Form: To apply the perpendicular distance formula, we need to express Equation 1 in the general linear form $Ax + By + C = 0$: $y = mx + \frac{1}{4m}$ $mx - y + \frac{1}{4m} = 0$ Here, $A = m$, $B = -1$, and $C = \frac{1}{4m}$.
• Apply Distance Formula: For the line to be tangent to the circle, the perpendicular distance from the center $(0,0)$ to the line $mx - y + \frac{1}{4m} = 0$ must be equal to the radius $\sqrt{2}$. $d = \frac{|A(0) + B(0) + C|}{\sqrt{A^2 + B^2}} = R$ $\frac{\left|m(0) - 1(0) + \frac{1}{4m}\right|}{\sqrt{m^2 + (-1)^2}} = \sqrt{2}$ $\frac{\left|\frac{1}{4m}\right|}{\sqrt{m^2 + 1}} = \sqrt{2}$ Why this step? This is the crucial step where we enforce the condition that the tangent line to the parabola must also be tangent to the circle. This will give us an algebraic equation involving $m$ that, when solved, will yield the slopes of the common tangents.

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Step 3: Solve for the Slopes ($m$)

• Simplify and Solve for m: To eliminate the absolute value and square root, we square both sides of the equation: $\left(\frac{\frac{1}{4m}}{\sqrt{m^2 + 1}}\right)^2 = (\sqrt{2})^2$ $\frac{\left(\frac{1}{4m}\right)^2}{m^2 + 1} = 2$ $\frac{\frac{1}{16m^2}}{m^2 + 1} = 2$ $\frac{1}{16m^2(m^2 + 1)} = 2$ Now, multiply both sides by $16m^2(m^2 + 1)$: $1 = 32m^2(m^2 + 1)$ $1 = 32m^4 + 32m^2$ Rearrange this into a standard quadratic form (in terms of $m^2$): $32m^4 + 32m^2 - 1 = 0$ Why this step? This polynomial in $m$ (specifically, a quadratic in $m^2$) will give us the values of $m$ that satisfy both tangency conditions. The problem states there are two common tangents with slopes $m_1$ and $m_2$.

• Solving the Quadratic in $m^2$: Let $M = m^2$. The equation becomes: $32M^2 + 32M - 1 = 0$ We use the quadratic formula $M = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to solve for $M$: $M = \frac{-32 \pm \sqrt{(32)^2 - 4(32)(-1)}}{2(32)}$ $M = \frac{-32 \pm \sqrt{1024 + 128}}{64}$ $M = \frac{-32 \pm \sqrt{1152}}{64}$ To simplify $\sqrt{1152}$: $1152 = 576 \times 2 = 24^2 \times 2$. So, $\sqrt{1152} = 24\sqrt{2}$. $M = \frac{-32 \pm 24\sqrt{2}}{64}$ Divide the numerator and denominator by 8: $M = \frac{-4 \pm 3\sqrt{2}}{8}$ This gives two possible values for $M$: $M_1 = \frac{-4 + 3\sqrt{2}}{8}$ $M_2 = \frac{-4 - 3\sqrt{2}}{8}$ Why we need to check $M$ values: Remember that $M = m^2$. For $m$ to be a real slope, $m^2$ must be non-negative ($m^2 \ge 0$). Let's approximate $3\sqrt{2} \approx 3 \times 1.414 = 4.242$. For $M_1$: $M_1 = \frac{-4 + 4.242}{8} = \frac{0.242}{8} > 0$. This is a valid value for $m^2$. For $M_2$: $M_2 = \frac{-4 - 4.242}{8} = \frac{-8.2