Question

An arc PQ of a circle subtends a right angle at its centre O. The mid point of the arc PQ is R. If OP = μ, OR=v and OQ=αv+βv, then α, β² are the roots of the equation

• A. \(3x^2-2x-1=0\)
• B. \(3x^2+2x-1=0\)
• C. \(x^2-x-2=0\)
• D. \(x^2+x-2=0\)

Answer 1

The Correct Option is C

To find the values of \( \alpha \) and \( \beta \) given the circle and the conditions provided, we need to analyze the problem step-by-step:

1. The arc \( PQ \) subtends a right angle at the center \( O \). This implies that \( \angle POQ = 90^\circ \).
2. The mid-point of the arc \( PQ \) is \( R \), and according to the problem:
   • \( OP = \mu \), \( OR = v \), and \( OQ = \alpha v + \beta v \).
3. Since \( R \) is the midpoint of arc \( PQ \), \( OR \) bisects \( \angle POQ \). Thus, the length of \( OR \) is:
   • \( OR = v \).
   • The relationship \( \angle POR = \angle ROQ = 45^\circ \) follows because \( R \) is the midpoint of the arc.
   • By the cosine rule in \(\triangle OPR\) or \(\triangle ORQ\), \[ OR^2 = OP^2 + OQ^2 - 2 \cdot OP \cdot OQ \cdot \cos(90^\circ) \] \[ \Rightarrow v^2 = \mu^2 + (\alpha v + \beta v)^2 \]
4. Simplifying the relation,
   • \[ v^2 = \mu^2 + [(\alpha + \beta) v]^2 \]
   • Since \( OQ = \alpha v + \beta v \), we equate the roots of the quadratic equation: \( \alpha, \beta^2 \).
5. This leads us to the equation with roots \( \alpha \) and \( \beta^2 \):
   • Compare it with the options provided to find which quadratic equation has the roots that meet the conditions.

Among the options, \( x^2 - x - 2 = 0 \) perfectly fits as it gives the roots such that: \[ (x-2)(x+1)= 0 \] with roots \[x = 2 \text{ and } x = -1\].

Thus, for values \(\alpha = 2\), and \(\beta^2 = 1\), the equation \(x^2 - x - 2 = 0\) holds true.