Multiple choice

Two concentric circles with centre $C$ have radii ${r}{1}$ and ${r}{2}$ such that $\left( { r }{ 2 }-{ r }{ 1 } \right) >0$. $CA$ and $CB$ are the common lined radii of the circles. If tangent at $A$ is drawn to meet the bigger circle in the point $D$. Then the length $BD$ is given by

  1. $\sqrt { 2{ r }_{ 1 }\left( { r }_{ 2 }-{ r }_{ 1 } \right) } $
  2. $\sqrt { 2{ r }_{ 2 }\left( { r }_{ 2 }-{ r }_{ 1 } \right) } $
  3. $\sqrt { 2{ r }_{ 1 }\left( { r }_{ 2 }+{ r }_{ 1 } \right) } $
  4. $\sqrt { 2{ r }_{ 2 }\left( { r }_{ 2 }+{ r }_{ 1 } \right) } $
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B Correct answer
Explanation

In the concentric circles, triangle CAD is a right triangle with hypotenuse r2 and leg r1. AD^2 = r2^2 - r1^2. In triangle CBD, CB = r1, CD = r2, and angle BCD is the angle between radii. Using the law of cosines or coordinate geometry, BD is calculated as sqrt(2*r2*(r2-r1)).

AI explanation

Let the centre C, point A on the smaller circle, and point B on the larger circle lie along the same line, making CB equal to r2. The tangent at A intersects the larger circle at D, forming right triangle CAD with the right angle at A. Since CA equals r1 and CD equals r2, we apply the geometric property of a tangent from an external point to find BD. In triangle CBD, we have CB equals r2 and CD equals r2, but correcting the setup using the right triangle where AD squared equals CD squared minus CA squared gives AD squared equals r2 squared minus r1 squared. Using the alternate segment theorem or chord properties, the length BD squared equals 2 times r2 times the difference (r2 minus r1), so BD equals the square root of 2 times r2 times (r2 minus r1).

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