Circles
Circle
star_batch_jee_advanced_2025
Grade 11
Question:
$ABC$ is a right angled triangle right angled at $A$, with side $AC = 1$ and $AB = a$, a circle having $AC$ as diameter cuts the side $CB$ at $D$ if $CD = b$ then:
ab > 1
ab < 1
ab > 1
\frac{1}{a} + \frac{1}{b} + 1 = \sqrt{\frac{1}{a^2} + \frac{1}{b^2}}
Step-by-Step Solution
Key Concept: Apply the constraint $\tan C = a$ to establish relationships between sides and derive the inequality for $\frac{b}{a}$ using the Pythagorean theorem.
In triangle $ADC$, we have $\tan C = \frac{AD}{b} = \frac{a}{1}$, which gives $AD = ab$ and $AC = 1$, so $ab a^2 + \frac{1}{2}$.
Correct Answer: 2,3