Problem 29 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(P Q\) and \(R S\) be tangents at the extremeties 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 (A) \(\sqrt{P Q \cdot R S}\) (B) \(\frac{P Q+R S}{2}\) (C) \(\frac{2 P Q \cdot R S}{P Q+R S}\) (D) \(\sqrt{\frac{P Q^{2}+R S^{2}}{2}}\)

Step-by-Step Solution

Answer

(A) \( \sqrt{P Q \cdot R S} \)

Step 1: Identify the Circle Properties

We know that the line segments \(PQ\) and \(RS\) are tangents to the circle at the extremities of the diameter \(PR\). According to the tangent-segment theorem, tangents drawn from the same external point are equal. Hence, we have \(PQ = RQ\) and \(PS = RS\).

Step 2: Understand Intersection and Circle Geometry

The points \(P\), \(R\), and \(X\) are points on the circle. Given that line \(PS\) and \(RQ\) intersect at \(X\) which is on the circumference, \(PXQRS\) is a cyclic quadrilateral.

Step 3: Summarize Properties of the Quadrilateral

In a cyclic quadrilateral, opposite angles sum to \(180^\circ\). Additionally, since \(P\) and \(R\) are endpoints of the diameter, \(\angle PXR = 90^\circ\). This means \(PQRS\) is a rectangle as each angle is \(90^\circ\).

Step 4: Use the Property of Equal Lengths and Right Angles

In rectangle \(PQRS\), the diagonals are equal and bisect each other. Therefore \(PQ = RS\) and the tangents are equal, reinforcing \(PQ = RQ\) and \(PS = RS\).

Step 5: Evaluate Possible Answers Using Geometric Interpretation

Given \(PQRS\) is a rectangle and \(X\) lies on the circle which implies its position equidistant from points \(P\) and \(R\), the length \(PR = 2r\) matches with the formulas. Check each option, only \( (A) \) fits such that \(2r = \sqrt{PQ \cdot RS}\) due to the properties of tangents and their equal distances from \(X\).

Step 6: Conclude the Correct Answer

Through geometric properties and expressions we determine: \(2r = \sqrt{PQ \cdot RS}\) since corresponding tangent segments from external points are equal.

Key Concepts

Cyclic Quadrilateral

Consider a quadrilateral formed inside a circle, known as a cyclic quadrilateral. For such quadrilaterals, all four vertices lie on the circumference of a circle. A remarkable property of cyclic quadrilaterals is that the sum of their opposite angles is always 180 degrees. This property can be observed in our figure when line segments intersect within the circle’s boundary, ultimately forming a cyclic quadrilateral.

In the geometry problem we have:

• Points P, R, and X lie on the circle, and
• Lines PS and RQ intersect at X, making PQRS a cyclic quadrilateral.

This valuable geometric feature is key for solving many circle-based problems, including the angles and tangents drawn among them. Recognizing a quadrilateral as cyclic can simplify calculations and solutions significantly.

Tangent-Segment Theorem

The tangent-segment theorem is a crucial rule when discussing tangents to a circle. When two tangents are drawn from an external point, they are equal in length. This theorem helps greatly in computations regarding circle geometry as it provides an avenue of symmetry and predictable consistency.

From our exercise, the external point provides tangents PQ and RS, which by the tangent-segment theorem, have equal lengths. This equality is key in simplifying our operations and understanding the relationship between different segments. Here, the equality was utilized to establish the dimensions of rectangle PQRS, reinforcing the cyclic quadrilateral properties.

• Helps to identify that PQ = RS
• Provides geometric insight to deduce equal segment lengths forming the figure

Recognizing and utilizing these equalities is a powerful method for solving complex problems in circle geometry.

Rectangle Properties

In this problem, once we establish a quadrilateral is cyclic, another level of geometry unravels when it becomes a rectangle. A rectangle possesses several distinguishing features:

• All angles are right angles (90°).
• Opposite sides are not only equal, but the diagonals also bisect each other at equal lengths.

Within our geometric figure, PQRS is deduced to be a rectangle because P and R are endpoints of the circle's diameter, inherently making \(angle PXR = 90°\).

These properties directly lead us to understand how diagonals PR and intersections maintain the integrity of equal lengths and positions, crucial for deriving the conditions and final answers in the exercise. Recognizing the form of quadrilaterals can dramatically ease reaching solutions.

Diameter

The diameter of a circle is the longest chord passing through the center of the circle. It directly affects understanding circle symmetry and some intensive properties. It is twice the radius, which gives it key geometrical value in calculations.

• The diameter helps establish the line's role and position within the circle, forming natural bisectors and dividing the circle into two equal halves.
• It can serve as a baseline for tangent construction as well, aiding visual clarity and calculative deduction.

In our exercise, noticing P and R as endpoints of the diameter gives us critical insights. Linking these endpoints via a diameter ensures a right angle at any third point X on the circle, due to the angle in a semicircle theorem, revealing more about tangent and chord relationships.

The concept of diameter is central, providing a geometric backbone into resolving exercises with precision and accuracy.