Matching Tangent and Point of Contact for Circle with Radius Root Two

To determine the correct combination for the conic, equation of tangent, and point of contact, we analyze the options given in the columns.

1. Identifying the Conic (Column 1)

Given the parameter $a = \sqrt{2}$, we test the given point of contact $(-1, 1)$ on the conics listed in Column 1:

• Conic (I): $x^2 + y^2 = a^2$ Substituting $x = -1$, $y = 1$, and $a = \sqrt{2}$: $(-1)^2 + (1)^2 = 1 + 1 = 2 = (\sqrt{2})^2 = a^2$ Since the equation is satisfied, the point $(-1, 1)$ lies on the circle $x^2 + y^2 = a^2$.

• Conic (II): $x^2 + a^2 y^2 = a^2 \implies x^2 + 2y^2 = 2$ $(-1)^2 + 2(1)^2 = 3 \neq 2$ Hence, the point does not lie on conic (II).

• Conic (III): $y^2 = 4ax = 4\sqrt{2}x$ $(1)^2 = 1 \neq 4\sqrt{2}(-1)$ Hence, the point does not lie on conic (III).

Therefore, the appropriate conic is (I).

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2. Identifying the Equation of Tangent (Column 2)

The standard equation of a tangent to the circle $x^2 + y^2 = a^2$ with slope $m$ is given by: $y = mx + a\sqrt{m^2 + 1}$

This matches item (ii) in Column 2.

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3. Identifying the Point of Contact (Column 3)

To express the point of contact $(x_1, y_1)$ in terms of the slope $m$, we compare the tangent equation in slope form $mx - y + a\sqrt{m^2 + 1} = 0$ with the equation of tangent at $(x_1, y_1)$, which is $x x_1 + y y_1 - a^2 = 0$:

$\frac{x_1}{m} = \frac{y_1}{-1} = \frac{-a^2}{a\sqrt{m^2 + 1}}$

Solving for $x_1$ and $y_1$: $x_1 = \frac{-ma}{\sqrt{m^2 + 1}}, \quad y_1 = \frac{a}{\sqrt{m^2 + 1}}$

Thus, the point of contact in terms of $m$ is: $\left(\frac{-ma}{\sqrt{m^2 + 1}}, \frac{a}{\sqrt{m^2 + 1}}\right)$

This matches item (Q) in Column 3.

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4. Verification for $a = \sqrt{2}$ and Point $(-1, 1)$

Equating the general point of contact $(Q)$ to $(-1, 1)$ with $a = \sqrt{2}$: $\frac{-m\sqrt{2}}{\sqrt{m^2 + 1}} = -1 \quad \text{and} \quad \frac{\sqrt{2}}{\sqrt{m^2 + 1}} = 1$

From the second equation, $\sqrt{m^2 + 1} = \sqrt{2} \implies m^2 + 1 = 2 \implies m = \pm 1$.
Substituting $m = 1$ into the first equation yields $\frac{-\sqrt{2}}{\sqrt{2}} = -1$, which is consistent.

Hence, the combination (I) (ii) (Q) is correct.

Correct Option: B