Let PQ and RS be tangents at the extremities of the diameter PR of a circle of radius r. If PS and R

Q: Why did we choose the y-axis for the diameter? Does it matter?

It is a choice of convenience! Coordinate geometry is invariant under rotation. Placing the diameter on the y-axis makes the tangents horizontal, which simplifies the equations significantly. The geometry remains the same regardless of orientation, but your algebraic workload decreases drastically with a smart setup.

Q: Is there a faster way using similar triangles?

Absolutely! In JEE Advanced, recognizing similar triangles is a superpower. You could prove $\triangle PSX \sim \triangle XQR$ or similar relationships. However, the coordinate geometry approach is a robust, systematic method that guarantees an answer even if you don't spot the geometric similarity immediately. It is a 'fail-safe' method.

Q: Why does the algebra look so messy before it simplifies?

That is the 'JEE Advanced Filter.' The problem is designed to test your patience and algebraic manipulation skills. Trust the process. When you see terms like $(x_1 + x_2)^2$ and $(x_2 - x_1)^2$ appearing together, it is a signal that an identity like $(a+b)^2 - (a-b)^2 = 4ab$ is waiting to save you. Keep pushing!

Question

Let $PQ$ and $RS$ be tangents at the extremities of the diameter $PR$ of a circle of radius $r$. If $PS$ and $RQ$ intersect at a point $X$ on the circumference of the circle, then $2r$ equals

Options:

(A) $\sqrt{PQ \cdot RS}$

(B) $(PQ + RS)/2$

(C) $2PQ \cdot RS/(PQ + RS)$

(D) $\sqrt{(PQ^2 + RS^2)/2}$

Sigma Percentile · Accepted Answer

### The Geometry of Elegance: Unlocking the Circle

Welcome, future engineer. Today, we are not just solving a problem; we are peeling back the layers of a geometric puzzle.

When you look at a problem involving tangents and circles, your first instinct might be to panic at the complexity. But in JEE Advanced, the most complex-looking problems often hide the most elegant, symmetric solutions. Let us walk through this journey together.

### Phase 1: The Canvas of Coordinates

Imagine you are standing before a blank whiteboard. We have a circle of radius $r$.

The most powerful tool we have is the ability to define our own reality. By placing the center of the circle at the origin $(0,0)$, we define the circle by the pristine equation:
$$x^2 + y^2 = r^2$$

We align the diameter $PR$ along the y-axis. This is a strategic move. Point $P$ becomes $(0, r)$ and point $R$ becomes $(0, -r)$.

Why do this? Because the tangents at the extremities of a diameter are parallel. By aligning the diameter vertically, our tangents become horizontal lines: $y = r$ and $y = -r$. We have just turned a complex geometric interaction into a simple linear system.

### Phase 2: The Lines of Engagement

Now, consider points $Q$ and $S$. $Q$ sits on the top tangent, so its coordinates are $(x_1, r)$. $S$ sits on the bottom tangent, so its coordinates are $(x_2, -r)$.

The lengths $PQ$ and $RS$ are simply $x_1$ and $x_2$. We need to find the intersection of lines $PS$ and $RQ$.

Let us find the equation of line $PS$. It passes through $(0, r)$ and $(x_2, -r)$. The slope is:
$$m = \frac{-r - r}{x_2 - 0} = -\frac{2r}{x_2}$$

Using the point-slope form, we get:
$$y - r = -\frac{2r}{x_2}x$$

Similarly, for line $RQ$, passing through $(0, -r)$ and $(x_1, r)$, the slope is:
$$m = \frac{r - (-r)}{x_1 - 0} = \frac{2r}{x_1}$$

The equation becomes:
$$y + r = \frac{2r}{x_1}x$$

### Phase 3: The Intersection

We are looking for the point $X(x, y)$ where these two lines meet. This is where the algebra begins to dance.

We have two expressions for $y$:
$$y = r - \frac{2rx}{x_2} \quad 	ext{and} \quad y = -r + \frac{2rx}{x_1}$$

Equating them gives us:
$$r - \frac{2rx}{x_2} = -r + \frac{2rx}{x_1}$$

Rearranging this, we find:
$$2r = 2rx\left(\frac{1}{x_1} + \frac{1}{x_2}ight)$$

The $2r$ cancels out beautifully, leaving us with $1 = x\left(\frac{x_1 + x_2}{x_1 x_2}ight)$. Thus, the x-coordinate of our intersection point is:
$$x = \frac{x_1 x_2}{x_1 + x_2}$$

### Phase 4: The Circle's Embrace

We are told that $X$ lies on the circumference. This is the constraint that binds the system.

We substitute our $x$ and the corresponding $y$ into $x^2 + y^2 = r^2$. Substituting $x = \frac{x_1 x_2}{x_1 + x_2}$ and $y = r\left(\frac{x_2 - x_1}{x_1 + x_2}ight)$, we get:
$$\left(\frac{x_1 x_2}{x_1 + x_2}ight)^2 + \left(r \frac{x_2 - x_1}{x_1 + x_2}ight)^2 = r^2$$

This looks like a mountain of algebra, but look closely. If we move the $y^2$ term to the right, we get:
$$\left(\frac{x_1 x_2}{x_1 + x_2}ight)^2 = r^2 \left[1 - \left(\frac{x_2 - x_1}{x_1 + x_2}ight)^2ight]$$

### Phase 5: The Algebraic Symphony

Focus on the bracket:
$$1 - \left(\frac{x_2 - x_1}{x_1 + x_2}ight)^2 = \frac{(x_1 + x_2)^2 - (x_2 - x_1)^2}{(x_1 + x_2)^2}$$

Here is the magic. The numerator is $(x_1 + x_2)^2 - (x_2 - x_1)^2$. Using the identity $(a+b)^2 - (a-b)^2 = 4ab$, this simplifies to $4x_1 x_2$.

So, our equation becomes:
$$\frac{(x_1 x_2)^2}{(x_1 + x_2)^2} = r^2 \frac{4 x_1 x_2}{(x_1 + x_2)^2}$$

The denominators cancel out! We are left with $(x_1 x_2)^2 = 4 r^2 x_1 x_2$. Dividing by $x_1 x_2$, we get:
$$x_1 x_2 = 4r^2$$

Since $x_1 = PQ$ and $x_2 = RS$, we have $PQ \cdot RS = 4r^2$. Taking the square root, we find:
$$2r = \sqrt{PQ \cdot RS}$$

We have arrived. The diameter is the geometric mean of the tangent segments. It is not just an answer; it is a testament to the harmony of mathematics.

Animated Solution for Mathematics - Circles: Let $P Q$ and $R S$ be tangents at the extremities of the diameter $P R$ of a circle of radius $r$ . If $P S$ and $R Q$ intersect at a point $X$ on the circumference of the circle, then $2 r$ equals

Select Answer:

Visualized Solution

Setting up the Coordinate System

Tangents and Points $Q$ , $S$

Equation of Line $P S$

Equation of Line $R Q$

Intersection Point $X$

Solving for $x$ -coordinate

Solving for $y$ -coordinate

The Circle Constraint

Substituting Coordinates

Algebraic Simplification

Expanding the Right Side

Equating Numerators

Final Result

The Sigma Insight: Equation of Tangent and Normal

The Geometry of Elegance

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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