Solved Examples - Circles Connected with Triangles | Mock Tests for JEE Main and Advanced 2025 PDF Download

Illustration 1: A straight line through the vertex P of a triangle PQR intersects the side QR at the point S and the circum circle of the triangle PQR at the point T. If S is not the center of the circum circle then which of the following relation is true? (2008)
1. 1/PS + 1/ST < 2/√QS.SR
2. 1/PS + 1/ST > 2/√QS.SR
3. 1/PS + 1/ST < 4/QR
4. 1/PS + 1/ST > 4/QR
Solution: Let us suppose that a straight line through the vertex P of a given ?PQR intersects the side QR at the point S and the circum circle of the triangle PQR at the point T. Now, the points P, Q, R and T are concyclic and hence PS.ST = QS.SR
Now we know that A.M > G.M
So, we have, (PS + ST)/ 2 > √PS.ST
1/ PS + 1/ ST > 2/ √PS.ST = 2/ √QS.SR
Also, (SQ + QR)/ 2 > √QS.SR
So, QR/2 > √QS.SR
1/√QS.SR > 2/QR
Hence, 2/ √QS.SR > 4/ QR
So, we obtain 1/ PS + 1/ ST > 2/ √QS.SR > 4/ QR.

Illustration 2: If in a triangle ABC, (2 cos A)/a + (cos B)/b + (2 cos C)/c = a/bc + b/ca. Find the value of angle A. (1993)
Solution: The given condition is (2 cos A)/a + (cos B)/b + (2 cos C)/c = a/bc + b/ca.     ….. (1)
We know that in ?ABC, by cosine rule we have
cos A = b² + c² - a² /2bc
and so, cos B = c² + a² - b² /2ac
and cos C = a² + b² - c² /2ab
We now substitute these values in equation (1) and hence obtain,
[2 (b² + c² - a²)/ 2abc] + [(c² + a² - b²) /2abc] + [2(a² + b² - c²)/ 2abc] = a/bc + b/ca
This gives [2(b² + c² - a²) + (c² + a² - b²) + 2(a² + b² - c²)] / abc
= (a² + b²)/abc
This means 3b² + c² + a² = 2a² + 2b²
Hence, b² + c² = a².
Hence we get the measure of angle A is 90°.

Illustration 3: In is the area of n-sided regular polygon inscribed in a circle of unit radius and On be the area of the polygon circumscribing the given circle, then prove that
I_n = O_n/2 (1 + √{1 – (2I_n/n)²}) (2003)
Solution: The given circle is of unit radius. An n-sided polygon of area In is inscribed inside a circle and another polygon of area On is circumscribed around the given circle.
We know that I_n = nr²/2 .sin 2π/n
Hence, since r = 1, so 2I_n / n = sin 2π/n    ….. (1)
Also, O_n = nr² tan π/2  
Hence, O_n/ n = tan π/n
Therefore 2I_n / O_n =   (sin 2π/n) / (tan π/n)
Hence, I_n / O_n = cos2π/n
= (1 + cos 2π/n)/2
Therefore, from equation (1) we obtain,
I_n / O_n = {1 + √[1 – (2I_n/ n)²]}/ 2
Hence, I_n = (O_n/ 2). (1 + √[1 – (2I_n/ n)²]
Hence proved.