Animated Solution for Mathematics - Circles: Comprehension Passage

Unlocking the Inscribed Circle

Geometry is not just about shapes on a page; it is about the hidden relationships that govern space. When we look at an equilateral triangle with an inscribed circle, we are looking at a masterpiece of symmetry.

Today, we are going to peel back the layers of this problem, transforming a seemingly complex set of coordinates into a clear, logical journey.

The Hunt for the Center

We start with a line $PQ$ given by $\sqrt{3}x+y-6=0$ and a point of contact $D$ at $(\frac{3\sqrt{3}}{2},\frac{3}{2})$. Our first goal is to find the center $C$ of the circle.

The most important geometric rule here is that the radius is perpendicular to the tangent. This means the center $C$ must lie on the normal line passing through $D$.

The slope of $PQ$ is $-\sqrt{3}$. Therefore, the slope of our normal line $CD$ is the negative reciprocal, $\frac{1}{\sqrt{3}}$.

This slope corresponds to an angle of $30^{\circ }$ with the positive x-axis. Using the parametric form of a line, we can step away from $D$ by the radius $r=1$.

The coordinates of the center are given by $C=(x_D\pm r\cos 30^{\circ },y_D\pm r\sin 30^{\circ })$. Substituting our values, we get two potential centers:

The Origin Constraint

Now, we face a fork in the road. Which center is the correct one? The problem provides a crucial constraint: the origin and the center $C$ must lie on the same side of the line $PQ$.

Let's define the expression $L(x,y)=\sqrt{3}x+y-6$. Testing the origin $(0,0)$, we get $L(0,0)=-6$, which is negative.

Now, we test our candidates: For $C_1(2\sqrt{3},2)$, we get $L(C_1)=2$, which is positive. For $C_2(\sqrt{3},1)$, we get $L(C_2)=-2$, which is negative.

Since $C_2$ shares the same sign as the origin, we have found our true center: $C=(\sqrt{3},1)$. With the center and radius known, the equation of the circle is:

The Elegance of Equilateral Symmetry

With the circle defined, we turn to the triangle $PQR$. In an equilateral triangle, the incenter is the centroid. This is a gift!

It means the contact points $D,E,F$ are the midpoints of the sides. The distance from the midpoint $D$ to the vertices $P$ and $Q$ is $r\tan 60^{\circ }=\sqrt{3}$.

Since the line $PQ$ has a slope of $-\sqrt{3}$, which is $120^{\circ }$, we can find $P$ and $Q$ by moving a distance of $\sqrt{3}$ from $D$ along the line $PQ$. This calculation yields:

The Final Coordinates

Finally, we find the third vertex $R$. Since $C$ is the centroid, we use the property:

Plugging in our known coordinates, we solve for $R$ and find it is at the origin $(0,0)$. With all vertices known, finding the contact points $E$ and $F$ is trivial: they are the midpoints of $QR$ and $PR$.

$E$ becomes $(\frac{\sqrt{3}}{2},\frac{3}{2})$ and $F$ becomes $(\sqrt{3},0)$. The equations of the sides $QR$ and $RP$ follow directly from the two-point form, completing our journey.

We have navigated the geometry, respected the constraints, and arrived at the solution through the sheer power of symmetry. Keep practicing this mindset, and you will find that even the most daunting JEE problems are just puzzles waiting to be solved.