Multiple choice

If the chord of contact of the tangents from a point on the circle $x ^ { 2 } + y ^ { 2 } = a ^ { 2 }$ to the circle $x ^ { 2 } + y ^ { 2 } = b$ touches the circle $x ^ { 2 } + y ^ { 2 } = c ^ { 2 } ,$ then $a , b ,$ care in

  1. $A G P$
  2. $A . P $
  3. $G . P $
  4. $H . P $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a circle centered at the origin, the chord of contact from a point at distance a to a circle of radius b lies at distance b^2/a from the origin. If this chord touches the circle of radius c, then b^2/a = c, so b^2 = ac. Therefore, a, b, c are in geometric progression.

AI explanation

Let the point be $(x_1, y_1)$ on $x^2 + y^2 = a^2$, so $x_1^2 + y_1^2 = a^2$. The chord of contact to the circle $x^2 + y^2 = b$ is $xx_1 + yy_1 = b$. The perpendicular distance from the origin to this line is $|b|/\sqrt{x_1^2 + y_1^2}$, which simplifies to $b/a$. Since this line touches $x^2 + y^2 = c^2$, the distance equals the radius $c$, giving $b/a = c$ and therefore $b = ac$. This means $b^2 = ac$, proving that $a, b, c$ form a geometric progression (G.P.).

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