Question

There are two circles in xy-plane whose equations are \( x^2 + y^2 - 2y = 0 \) and \( x^2 + y^2 - 2y - 3 = 0 \). A point (x, y) is chosen at random inside the larger circle. Then the probability that the point has been taken from smaller circle is

• A. \( \frac{1}{3} \)
• B. \( \frac{2}{3} \)
• C. \( \frac{1}{2} \)
• D. \( \frac{1}{4} \)

Hint:

In geometric probability problems, the probability of choosing a point from a sub-region S within a larger region L is simply the ratio of their measures (Area(S) / Area(L)). To find the area of a circle from its general equation, complete the square to find its radius.

Answer 1

The Correct Option is D

This is a geometric probability problem. The probability is the ratio of the favorable area to the total possible area. 

Probability = \( \frac{\text{Area of smaller circle}}{\text{Area of larger circle}} \). 

Let's find the centers and radii of the two circles by converting their equations to the standard form \( (x-h)^2 + (y-k)^2 = r^2 \). 

Circle 1 (Smaller Circle): \[ x^2 + y^2 - 2y = 0 \] Complete the square for the y-terms: \[ x^2 + (y^2 - 2y + 1) = 1 \] \[ x^2 + (y-1)^2 = 1^2 \] Center \( C_1 = (0, 1) \). Radius \( r_1 = 1 \). Area of smaller circle = \( \pi r_1^2 = \pi(1)^2 = \pi \). 

Circle 2 (Larger Circle): \[ x^2 + y^2 - 2y - 3 = 0 \] Complete the square: \[ x^2 + (y^2 - 2y + 1) = 3 + 1 \] \[ x^2 + (y-1)^2 = 4 = 2^2 \] Center \( C_2 = (0, 1) \). Radius \( r_2 = 2 \). Area of larger circle = \( \pi r_2^2 = \pi(2)^2 = 4\pi \). The smaller circle is entirely contained within the larger circle since they have the same center. Now, calculate the probability: \[ \text{Probability} = \frac{\text{Area of smaller circle}}{\text{Area of larger circle}} = \frac{\pi}{4\pi} = \frac{1}{4} \]