IIT-JEE 2008 · Paper 1 · Mathematics Q15

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MathematicsSingle correct+4 / -1Question 15 of 69

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Paragraph for Question Nos. 15 to 17

A circle $C$ of radius 1 is inscribed in an equilateral triangle $PQR$ . The points of contact of $C$ with the sides $PQ$ , $QR$ , $RP$ are $D$ , $E$ , $F$ , respectively. The line $PQ$ is given by the equation $\sqrt3x+y-6=0$ and the point $D$ is $\left(\frac{3\sqrt3}2,\frac32\right)$ . 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

(x-2\sqrt3)^2+(y-1)^2=1

(x-2\sqrt3)^2+\left(y+\frac12\right)^2=1

(x-\sqrt3)^2+(y+1)^2=1

(x-\sqrt3)^2+(y-1)^2=1

Answer and worked solution

Official answer: D

Centre. It lies on the normal to PQ through D, at distance 1, on the origin's side. The unit normal to $\sqrt3x+y-6=0$ is $\left(\frac{\sqrt3}2,\frac12\right)$ , pointing away from the origin (where the expression is negative). So
$\text{centre}=\left(\frac{3\sqrt3}2,\frac32\right)-\left(\frac{\sqrt3}2,\frac12\right)=(\sqrt3,1).$

Circle. $(x-\sqrt3)^2+(y-1)^2=1$ .

(D)

Topics:Equation of a circle Distance of a point from a line

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IIT-JEE 2008 · Paper 1, Mathematics question 15