UPSC CDS Math Circle Questions

P is the mid-point of QR.∴ ∠OPQ = 90°(Perpendicular from the centre of the circle to the chord bisects the chord)PQ = PR = 1 cmAlso, M is the mid-point of CD.∴∠OMP = 90°(Perpendicular from the centre of the circle to the chord bisects the chord)In right ΔOPQ $OP=\sqrt{OQ^2-PQ^2}=\sqrt{7^2-1^2}$$=\sqrt{48}\,\text{cm}$In right ΔOMP$OM=\sqrt{OP^2-MP^2}=\sqrt{48-24}$$=\sqrt{24}$cm∴ OM = MP⇒ ∠OPM = ∠POM = 45°∴ ∠QPM = 90° - 45° = 45°⇒∠QPD = 180° - 45° = 135°(Linear pair)Hence, statement 1 is correct. QR and CD are two chords of the circle that intersect at P.∴ CP × PD = QP × PR ⇒mn =1So, m and n are the roots of the quadratic equation$x^2-10x+1=0$,as the product of the roots is 1.Hence, statement 2 is correct.Area of ΔOPR$=\frac{1}{2} \times PR \times OP$$=\frac{1}{2} \times 1 \times \sqrt{48} = 2\sqrt{3}\,\text{cm}^2$Area of ΔOMP$=\frac{1}{2} \times MP \times OM$$=\frac{1}{2} \times \sqrt{24} \times \sqrt{24} = 12\,\text{cm}^2$Therefore, ratio of the area of ΔOPR to the area of ΔOMP$=\frac{2\sqrt{3}}{12} = \frac{1}{2\sqrt{3}} = 1:2\sqrt{3}$Hence, statement 3 is incorrect.