The circle C1 : x² + y² = 3, with centre at O, intersects the parabola x² = 2y at the point P in the

Q: Wait, why do we reject the negative root for $y$ in the intersection step?

Excellent question! We are looking for the intersection point $P$ specifically in the first quadrant. In the first quadrant, both $x$ and $y$ must be positive. Solving $y^2 + 2y - 3 = 0$ gives $y=1$ and $y=-3$. Since $y=-3$ is outside our quadrant of interest, we discard it. Always keep an eye on the constraints given in the problem statement!

Q: Why are there two circles $C_2$ and $C_3$?

The math tells the truth! When we solve the distance equation $|k-3| = 6$, the absolute value naturally splits into two cases: $k-3=6$ and $k-3=-6$. This corresponds to the two possible positions for the centers on the $y$-axis that satisfy the tangency condition. It's not a coincidence; it's the geometry of the line intersecting the $y$-axis.

Q: Is there a faster way to find $R_2R_3$ without calculating coordinates?

Yes! Think of $R_2R_3$ as the projection of the segment $Q_2Q_3$ onto the tangent line. If you know the angle $\phi$ between the normal to the line and the $y$-axis, you can use the relation $R_2R_3 = Q_2Q_3 \sin(\phi)$. It saves you from finding the exact coordinates of $R_2$ and $R_3$!

Question

The circle $C_1 : x^2 + y^2 = 3$, with centre at $O$, intersects the parabola $x^2 = 2y$ at the point $P$ in the first quadrant. Let the tangent to the circle $C_1$ at $P$ touches other two circles $C_2$ and $C_3$ at $R_2$ and $R_3$, respectively. Suppose $C_2$ and $C_3$ have equal radii $2\sqrt{3}$ and centres $Q_2$ and $Q_3$, respectively. If $Q_2$ and $Q_3$ lie on the $y$-axis, then

Options:

(A) $Q_2 Q_3 = 12$

(B) $R_2 R_3 = 4\sqrt{6}$

(C) area of the triangle $\Delta OR_2 R_3$ is $6\sqrt{2}$

(D) area of the triangle $\Delta PQ_2 Q_3$ is $4\sqrt{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Intersection of Destinies

We begin with two curves: the circle $C_1: x^2 + y^2 = 3$ and the parabola $x^2 = 2y$. They meet at a point $P$ in the first quadrant.

To find $P$, we substitute the parabola's equation into the circle's:
$$2y + y^2 = 3$$
$$y^2 + 2y - 3 = 0$$

Factoring this quadratic gives us $(y+3)(y-1) = 0$. We have two roots, $y=1$ and $y=-3$.

Since we are in the first quadrant, the negative root is discarded. Thus, $y=1$ is our solution. Plugging this back into $x^2 = 2y$, we find $x^2 = 2$, so $x = \sqrt{2}$. Our anchor point is $P(\sqrt{2}, 1)$.

### The Tangent Bridge

Now, we draw a tangent to $C_1$ at $P$. Using the formula $xx_1 + yy_1 = a^2$, we substitute our point $(\sqrt{2}, 1)$ to obtain:
$$\sqrt{2}x + y = 3$$
$$\sqrt{2}x + y - 3 = 0$$

This line serves as the boundary for our subsequent geometric constructions.

### The Mystery Circles

We are given that $C_2$ and $C_3$ have radii $2\sqrt{3}$ and their centers $Q_2, Q_3$ lie on the $y$-axis. Let a center be $Q(0, k)$.

The distance from this center to our tangent line must equal the radius, $2\sqrt{3}$. Using the perpendicular distance formula:
$$\frac{|\sqrt{2}(0) + k - 3|}{\sqrt{(\sqrt{2})^2 + 1^2}} = 2\sqrt{3}$$

This simplifies to:
$$\frac{|k-3|}{\sqrt{3}} = 2\sqrt{3} \Rightarrow |k-3| = 6$$

Solving for $k$, we get $k-3 = 6$ (so $k=9$) and $k-3 = -6$ (so $k=-3$). Our centers are $Q_2(0, 9)$ and $Q_3(0, -3)$. The distance $Q_2Q_3$ is $9 - (-3) = 12$.

### The Geometry of Projections

We seek the length $R_2R_3$, where $R_2$ and $R_3$ are the feet of the perpendiculars from the centers to the tangent line. The segment $R_2R_3$ is the projection of $Q_2Q_3$ onto the tangent line.

If $\phi$ is the angle between the normal to the line and the $y$-axis, then $R_2R_3 = Q_2Q_3 \sin(\phi)$. The normal vector is $(\sqrt{2}, 1)$ and the $y$-axis vector is $(0, 1)$.

The cosine of the angle $\phi$ is:
$$\cos(\phi) = \frac{|(\sqrt{2}, 1) \cdot (0, 1)|}{\sqrt{2+1} \cdot 1} = \frac{1}{\sqrt{3}}$$

Thus, $\sin(\phi) = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}}$. The length is:
$$R_2R_3 = 12 	imes \sqrt{\frac{2}{3}} = 4\sqrt{6}$$

### The Final Area

Finally, we calculate the area of $\Delta OR_2R_3$. The base is $R_2R_3 = 4\sqrt{6}$.

The height is the perpendicular distance from the origin $O(0,0)$ to the tangent line $\sqrt{2}x + y - 3 = 0$:
$$h = \frac{|-3|}{\sqrt{(\sqrt{2})^2 + 1^2}} = \frac{3}{\sqrt{3}} = \sqrt{3}$$

The area is:
$$	ext{Area} = \frac{1}{2} 	imes 4\sqrt{6} 	imes \sqrt{3} = 2\sqrt{18} = \mathbf{6\sqrt{2}}$$

Select Answer:

Visualized Solution

Visualizing $C_{1}$ and $x^{2} = 2 y$

Finding Point $P$

Equation of Tangent at $P$

Finding Centers $Q_{2}$ and $Q_{3}$

Distance $Q_{2} Q_{3}$

Locating $R_{2}$ and $R_{3}$

Length of $R_{2} R_{3}$

Area of $Δ O R_{2} R_{3}$

Area of $Δ P Q_{2} Q_{3}$

Final Summary

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Intersection of Destinies

The Tangent Bridge

The Mystery Circles

The Geometry of Projections

The Final Area

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Select Answer:

Visualized Solution

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