JEE Challenger
More from Conic Sections
MathMedAdv 2008 • Paper 1

Contact Points of Incircle on Triangle Sides

Comprehension Passage

A circle CC of radius 1 is inscribed in an equilateral triangle PQRPQR. The points of contact of CC with the sides PQ,QR,RPPQ, QR, RP are D,E,FD, E, F, respectively. The line PQPQ is given by the equation 3x+y−6=0\sqrt{3} x+y-6=0 and the point DD is (332,32)\left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right). Further, it is given that the origin and the centre of CC are on the same side of the line PQPQ.

Points EE and FF are given by

Options

A

(32,32),(3,0)\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right), (\sqrt{3}, 0)

Correct
B

(32,12),(3,0)\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right), (\sqrt{3}, 0)

C

(32,32),(32,12)\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right), \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)

D

(32,32),(32,12)\left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right), \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)

Topics & Concepts

Conic SectionsCirclesStraight Lines

Step-by-Step Solution

To find the points of contact EE and FF, we first determine the centre and the equation of the inscribed circle CC.

  1. Find the Centre of the Circle CC:
    The side PQPQ is given by the linear equation:
    3x+y−6=0\sqrt{3}x + y - 6 = 0

The unit normal vector perpendicular to the line PQPQ directed away from the origin side is:
n^=(3(3)2+12,1(3)2+12)=(32,12)\hat{n} = \left(\frac{\sqrt{3}}{\sqrt{(\sqrt{3})^2 + 1^2}}, \frac{1}{\sqrt{(\sqrt{3})^2 + 1^2}}\right) = \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)

Evaluating the line expression at the origin (0,0)(0,0) gives:
3(0)+0−6=−6<0\sqrt{3}(0) + 0 - 6 = -6 < 0

Since the centre of the circle, C0(h,k)C_0(h, k), and the origin lie on the same side of the line PQPQ, the centre must be shifted inwards from the point of contact D(332,32)D\left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right) along the normal by the radius r=1r = 1:
C0=D−rn^=(332−32, 32−12)=(3,1)C_0 = D - r\hat{n} = \left(\frac{3\sqrt{3}}{2} - \frac{\sqrt{3}}{2}, \, \frac{3}{2} - \frac{1}{2}\right) = (\sqrt{3}, 1)

  1. Locate the Points of Contact EE and FF:
    Since the circle is inscribed in an equilateral triangle PQRPQR, the points of contact D,E,D, E, and FF divide the circle into three equal arcs of 120∘120^\circ each.

The position vector of DD relative to the centre C0C_0 is:
r⃗D=D−C0=(332−3, 32−1)=(32,12)\vec{r}_D = D - C_0 = \left(\frac{3\sqrt{3}}{2} - \sqrt{3}, \, \frac{3}{2} - 1\right) = \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)

In polar form relative to C0C_0, this vector has radius r=1r = 1 and an angle θ\theta:
cos⁡θ=32,sin⁡θ=12  ⟹  θ=30∘\cos\theta = \frac{\sqrt{3}}{2}, \quad \sin\theta = \frac{1}{2} \implies \theta = 30^\circ

The other two points of contact, EE and FF, are obtained by rotating r⃗D\vec{r}_D by +120∘+120^\circ and −120∘-120^\circ (or +240∘+240^\circ):

  • For the first point:
    θ1=30∘+120∘=150∘\theta_1 = 30^\circ + 120^\circ = 150^\circ
    The coordinates are:
    C0+(cos⁡150∘,sin⁡150∘)=(3,1)+(−32,12)=(32,32)C_0 + (\cos 150^\circ, \sin 150^\circ) = (\sqrt{3}, 1) + \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right) = \left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right)

  • For the second point:
    θ2=30∘−120∘=−90∘\theta_2 = 30^\circ - 120^\circ = -90^\circ
    The coordinates are:
    C0+(cos⁡(−90∘),sin⁡(−90∘))=(3,1)+(0,−1)=(3,0)C_0 + (\cos(-90^\circ), \sin(-90^\circ)) = (\sqrt{3}, 1) + (0, -1) = (\sqrt{3}, 0)

Thus, the points EE and FF are:
(32,32)and(3,0)\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right) \quad \text{and} \quad (\sqrt{3}, 0)

This corresponds to Option A.

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