Multiple choice

Three circles with radii $3$ cm, $4$ cm and $5$ cm touch each other externally. If $A$ is the point of intersection of tangents to these circles at !heir points of contact. then !he distance of A from !he points of contact is

  1. $\sqrt{3} $
  2. $2$
  3. $\sqrt{5}$
  4. $\sqrt{6}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the tangent-length relationships for three externally touching circles with radii 3, 4, and 5 cm, the required tangent distance evaluates to sqrt(5) cm. Therefore, option C matches the result.

AI explanation

The given circles with radii 3, 4, and 5 touch each other externally, so the distances between their centers are 7, 8, and 9. Using Heron's formula for the triangle formed by the centers, the semi-perimeter s = (7+8+9)/2 = 12, and the area is sqrt(12 * 5 * 4 * 3) = sqrt(720) = 12*sqrt(5). The point of intersection of the common external tangents is the external center of similitude, which serves as the excenter of the triangle formed by the centers. The distance from this excenter to a point of contact (which corresponds to a vertex of the contact triangle) relates to the exradius, but for three mutually tangent circles, the distance from the radical center (which coincides with the tangency intersection point) to a point of tangency is given by sqrt(r1*r2*r3 / (r1+r2+r3)). Calculating this gives sqrt((3*4*5) / (3+4+5)) = sqrt(60 / 12) = sqrt(5).