A circle C of radius 1 is inscribed in an equilateral triangle PQR. The points of contact of C with the sides P Q, Q R, R P are D, E, F, respectively. The line P Q is given by the equation √3 x+y-6=0 and the point D is ((3 √3/2), (3/2)). Further, it is given that the origin and the centre of C are on the same side of the line PQ. The equation of circle C is

Question

A circle C of radius 1 is inscribed in an equilateral triangle PQR. The points of contact of C with the sides P Q, Q R, R P are D, E, F, respectively. The line P Q is given by the equation √3 x+y-6=0 and the point D is ((3 √3/2), (3/2)). Further, it is given that the origin and the centre of C are on the same side of the line PQ. The equation of circle C is (A) (x-2 √3)2+(y-1)2=1 (B) (x-2 √3)2+(y+(1/2))2=1 (C) (x-√3)2+(y+1)2=1 (D) (x-√3)2+(y-1)2=1

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/d5b2eec7cadbb862259f425dabaac16e-.png" /> Let G(α, β) be the centre of C α=(3 √3/2)-1 ⋅ cos 30=√3 β=(3/2)-1 ⋅ sin 30=1 ∴ equation of C is (x-√3)2+(y-1)2=1 ∴ D is correct.

$(x - 23)^{2} + (y - 1)^{2} = 1$

$(x - 23)^{2} + (y + \frac{1}{2})^{2} = 1$

$(x - 3)^{2} + (y + 1)^{2} = 1$

$(x - 3)^{2} + (y - 1)^{2} = 1$

Let $G (α, β)$ be the centre of $C$
$α = \frac{3 3}{2} - 1 \cdot cos 30 = 3$
$β = \frac{3}{2} - 1 \cdot sin 30 = 1$
$∴$ equation of $C$ is $(x - 3)^{2} + (y - 1)^{2} = 1$
$∴ D$ is correct.

(x−23​)2+(y−1)2=1

(x−23​)2+(y+21​)2=1

(x−3​)2+(y+1)2=1

(x−3​)2+(y−1)2=1

Let G(α,β) be the centre of C α=233​​−1⋅cos30=3​ β=23​−1⋅sin30=1 ∴ equation of C is (x−3​)2+(y−1)2=1 ∴D is correct.

$(x - 23)^{2} + (y - 1)^{2} = 1$

$(x - 23)^{2} + (y + \frac{1}{2})^{2} = 1$

$(x - 3)^{2} + (y + 1)^{2} = 1$

$(x - 3)^{2} + (y - 1)^{2} = 1$

Let $G (α, β)$ be the centre of $C$
$α = \frac{3 3}{2} - 1 \cdot cos 30 = 3$
$β = \frac{3}{2} - 1 \cdot sin 30 = 1$
$∴$ equation of $C$ is $(x - 3)^{2} + (y - 1)^{2} = 1$
$∴ D$ is correct.