Circles
Circle
star_batch_jee_advanced_2025
Grade None
Question:
There are two two circles in a parallelogram. One of them of radius 3units is inscribed in the parallelogram, and the other touches two sides of the parallelogram and the first circle. The distance between the points of tangency which lie on the same side of the parallelogram is equal to 3 units.
The radius of the other circle is $\frac{3}{4}$ units
Area of the parallelogram is equal to $75/2$ units
Let $d_1, d_2$ denote the lengths of the diagonals of parallelogram, then the product $d_1.d_2$ is equal to $75$
Let $d_1, d_2$ denote the lengths of the diagonals of parallelogram, then the product $d_1.d_2$ is equal to $95$
Step-by-Step Solution
Key Concept: The inradius and circumradius relationship with the rhombus geometry determines all dimensions through trigonometric identities.
From $2\sqrt{Rr} = 3$, we get $r = \frac{3}{4}$. Using $\frac{r}{R} = \frac{x}{x+3} = \frac{1}{4}$ gives $x = 1$. With $\tan\frac{\theta}{2} = \frac{3}{4}$, we find $y = R\cot\left(\frac{\pi}{2} - \frac{\theta}{2}\right) = 3\tan\frac{\theta}{2} = \frac{9}{4}$. The rhombus $ABCD$ has side $4 + \frac{9}{4} = \frac{25}{4}$ with area $\frac{25}{4} \times \frac{25}{4} \times \sin\theta = \frac{625}{16} \times \frac{2 \cdot \frac{3}{4}}{1 + \frac{9}{16}}$, and total area involves integrating $d_1d_2 = 75$.
Correct Answer: 1,2,3