Multiple choice

The circles ${ S }{ 1 }$ with centre ${ C }{ 1 }\left( { a }{ 1 },{ b }{ 1 } \right) $ and radius ${ r }{ 1 }$ touches externally the circle ${ S }{ 2 }$ with centre ${ C }{ 2 }\left( { a }{ 2 },{ b }{ 2 } \right) $ and radius ${ r }{ 2 }$. If the tangent at their common point passes through the origin, then

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

For two circles touching externally, the distance between centers equals the sum of radii: (a1-a2)^2 + (b1-b2)^2 = (r1+r2)^2. The tangent at the common point passes through the origin. This geometric condition leads to the relation (a1^2 - a2^2) + (b1^2 - b2^2) = r1^2 - r2^2.

AI explanation

Let the common point of contact be (x, y). Since the tangent passes through the origin and is perpendicular to the radius at the point of contact, the radius from the origin to the point (x, y) forms a right angle with the tangent line. This makes the distances from the origin to the centers equal to sqrt(r_1^2 + x^2 + y^2) and sqrt(r_2^2 + x^2 + y^2), leading to the relation a_1^2 + b_1^2 = r_1^2 + x^2 + y^2 and a_2^2 + b_2^2 = r_2^2 + x^2 + y^2. By subtracting the second equation from the first, the (x^2 + y^2) terms cancel out, leaving (a_1^2 - a_2^2) + (b_1^2 - b_2^2) = r_1^2 - r_2^2.

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