Multiple choice

$C_1$ and $C_2$ are the two concentric circles with radii $r_1$ and ${r}{2}({r}{1}<{r}_{2})$ . If the tangent drawn from any point of $C_2$ to $C_1$ meet again $C_2$ at the ends of its diameter, then

  1. ${r}_{2}=2{r}_{1}$
  2. ${r}_{2^{=}}\sqrt{2}{r}_{1}$
  3. ${r}_{2}^{2}<2{r}_{1}^{2}$
  4. ${r}_{1}=2{r}_{2}$
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B Correct answer
AI explanation

The tangent to the inner circle and the two radii of the outer circle form a right-angled triangle at the point of tangency. Using the Pythagorean theorem, the square of the outer radius equals the square of the inner radius plus the square of the tangent length. Since the tangent forms the diameter of the larger circle, its length is 2 times r2, making r2 squared equal to r1 squared plus 4 times r2 squared, which simplifies to r2 equals the square root of 2 times r1.

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