Question

Three circles A, B and C have a common centre O. A is the inner circle, B is the middle circle, and C is the outer circle. P is a point on the outer circle C, and the radius OP cuts the inner circle at X and the middle circle at Y, such that \(OX = XY = YP\). The ratio of the area of the region between the inner and middle circles to the area of the region between the middle and outer circles is:

• A. \(\dfrac{1}{3}\)
• B. \(\dfrac{2}{5}\)
• C. \(\dfrac{3}{5}\)
• D. \(\dfrac{1}{5}\)

Hint:

Since OX = XY = YP, the three circles have radii in the ratio 1:2:3; use area = pi times radius squared for each ring and take the ratio.

Answer 1

The Correct Option is C

Step 1: Express the three radii as parts.
Because $OX = XY = YP$, the point $X$ lies at 1 part from the centre, $Y$ lies at 2 parts, and $P$ lies at 3 parts, taking one segment length as one "part". So the radii of the inner, middle and outer circles are in the ratio $1 : 2 : 3$.

Step 2: Recall that the area of a circle scales with the square of its radius.
Since area $= \pi r^2$, radii in ratio $1:2:3$ give areas in ratio $1^2 : 2^2 : 3^2 = 1 : 4 : 9$.

Step 3: Write the three circle areas using this ratio.
Let one area unit be $k$. Then:
area of inner circle $= k$
area of middle circle $= 4k$
area of outer circle $= 9k$

Step 4: Find the two ring areas from these values.
Ring between inner and middle circles $= 4k - k = 3k$.
Ring between middle and outer circles $= 9k - 4k = 5k$.

Step 5: Take the ratio asked in the question.
\[ \frac{\text{ring (inner-middle)}}{\text{ring (middle-outer)}} = \frac{3k}{5k} = \frac{3}{5} \]
Working purely with the squared-ratio numbers $1:4:9$, without writing out $\pi r^2$ for an actual radius, gives the same result as computing the real areas, since $k$ cancels out either way.

Final Answer:
The ratio of the two ring areas is $3:5$. \[ \boxed{\dfrac{3}{5}} \]