In the circle given below, let OA = 1 unit, OB = 13 unit and PQ || OB. Then, the area of the triangl

Q: Wait, the problem says $PQ \parallel OB$, but the diagram shows it perpendicular. Which one should I trust?

In competitive exams like JEE, diagrams are often the source of truth when text descriptions seem contradictory or ambiguous. The visual evidence of the right angle at $A$ is the key. If $PQ$ were parallel to $OB$, it would be a tangent or a secant line, but the geometry of the triangle $PQB$ would not be well-defined in the same way. Trust the visual geometry: $PQ$ is a vertical chord perpendicular to the diameter $OB$.

Q: Why can't I just use the equation of the circle $(x-h)^2 + y^2 = R^2$?

You absolutely can! That is a fantastic, rigorous approach. If you set the center at $(6.5, 0)$ and the radius as $6.5$, the equation is $(x - 6.5)^2 + y^2 = 6.5^2$. When you plug in $x = 1$ (the position of the chord), you get $(1 - 6.5)^2 + y^2 = 6.5^2$, which simplifies to $(-5.5)^2 + y^2 = 6.5^2$. This leads you to the exact same result as the Pythagorean theorem. Both methods are valid; choose the one that feels more intuitive to you!

Q: Why is the height of $\triangle PQB$ equal to $AB$?

Think about the definition of the height of a triangle. It is the perpendicular distance from a vertex to the opposite side (the base). Here, our base is the vertical chord $PQ$. The vertex $B$ lies on the x-axis. The perpendicular distance from $B$ to the vertical line $x = 1$ is simply the horizontal distance along the x-axis, which is the segment $AB$. Since $A$ is at $x = 1$ and $B$ is at $x = 13$, the distance is simply $13 - 1 = 12$.

Question

In the circle given below, let $OA = 1$ unit, $OB = 13$ unit and $PQ \parallel OB$. Then, the area of the triangle PQB (in square units) is:

Options:

(A) $26\sqrt{3}$

(B) $24\sqrt{2}$

(C) $24\sqrt{3}$

(D) $26\sqrt{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we aren't just solving a geometry problem; we are learning how to translate the language of shapes into the powerful, precise dialect of coordinate geometry.

When you look at a problem like this, it is easy to feel overwhelmed by the lines and the curves. But I want you to take a deep breath. Geometry is not about memorizing formulas; it is about finding the hidden symmetry in the chaos.

### The Coordinate Transformation

Imagine you are standing on a blank canvas. The problem gives us a circle with a diameter $OB$. The most powerful tool in our arsenal is the Cartesian coordinate system.

By placing point $O$ at the origin $(0, 0)$ and aligning the diameter $OB$ along the positive x-axis, we have effectively 'tamed' the circle. We are told $OA = 1$ and $OB = 13$. Because $O$ is at the origin, $A$ sits at $(1, 0)$ and $B$ sits at $(13, 0)$.

Since $OB$ is the diameter, the center $C$ must be the midpoint of $OB$. The midpoint of $0$ and $13$ is:

$$C = \left( \frac{0 + 13}{2}, 0 ight) = (6.5, 0)$$

The radius $R$ is $6.5$. The circle is defined by the equation:

$$(x - 6.5)^2 + y^2 = 6.5^2$$

We have turned a visual puzzle into an algebraic certainty.

### The Chord and the Pythagorean Bridge

Now, let's address the chord $PQ$. The diagram shows it passing through $A$ and standing vertically. The visual geometry tells us that $PQ$ is perpendicular to the diameter.

To find the length of $PQ$, we need to find the distance from the center $C$ to the chord. We drop a perpendicular from $C$ to the chord at point $A$. This creates a right-angled triangle, $	riangle CAP$.

The hypotenuse $CP$ is the radius of the circle, which is $6.5$. The base $AC$ is the distance between the center $(6.5, 0)$ and the point $A(1, 0)$, which is $|6.5 - 1| = 5.5$.

Now, we invoke the Pythagorean theorem:

$$CP^2 = AC^2 + PA^2$$

Substituting our values, we get:

$$6.5^2 = 5.5^2 + PA^2$$

Instead of doing heavy arithmetic, let's use the difference of squares:

$$PA^2 = 6.5^2 - 5.5^2 = (6.5 - 5.5)(6.5 + 5.5) = 1 	imes 12 = 12$$

Thus, $PA = \sqrt{12} = 2\sqrt{3}$. Since the perpendicular from the center bisects the chord, the total length of $PQ$ is:

$$PQ = 2 	imes PA = 4\sqrt{3}$$

### The Final Area

We are in the home stretch. We need the area of $	riangle PQB$. The formula for the area of a triangle is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.

We have our base $PQ = 4\sqrt{3}$. The height of the triangle, relative to this base, is the perpendicular distance from vertex $B$ to the line $PQ$. Since $PQ$ is the vertical line $x = 1$ and $B$ is at $x = 13$, the height is simply the horizontal distance $AB = 13 - 1 = 12$.

Now, we calculate:

$$	ext{Area} = \frac{1}{2} 	imes (4\sqrt{3}) 	imes 12$$

$$	ext{Area} = 2\sqrt{3} 	imes 12 = 24\sqrt{3}$$

Look at that! The complexity melts away, leaving behind a clean, elegant result. The final area is **$24\sqrt{3}$**.

Animated Solution for Mathematics - Circles: In the circle given below, let $O A = 1$ unit, $O B = 13$ unit and $P Q ∥ O B$ . Then, the area of the triangle PQB (in square units) is:

Select Answer:

Visualized Solution

Coordinate System Setup

Locating Points $A$ and $B$

The Circle and its Diameter

Finding the Center and Radius

The Chord $P Q$

Setting up the Right Triangle

Calculating $A C$

Applying Pythagoras

Solving for $P A$

Length of Chord $P Q$

Triangle $P QB$

Base and Height of $△ P QB$

Calculating Height $A B$

Final Area Calculation

Final Answer

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Coordinate Transformation

The Chord and the Pythagorean Bridge

The Final Area

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Animated Solution for Mathematics - Circles: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In the circle given below, let OA=1 unit, OB=13 unit and PQ∥OB. Then, the area of the triangle PQB (in square units) is: