Question

Circle with radii 3, 4 and 5 touch each other externally, if P is the point of intersection of tangents to these circles at their points of contact. Find the distance of P from the point of contact.

Answer 1

In triangles Δ CP₃ & Δ APC₃.

CP = AP (Tangent on circle form point P)

PC₃ = PC₃ (Common side)

CC₃ = AC₃ = radii

\(\therefore \) Δ CPC₃ ≅ Δ APC₃

\(\therefore \) ∠CC₃P = ∠AC₃P

Similarly, ∠BGP = ∠AC₁P

∠BC₂P = ∠CC₂P

Hence, P is the point of intersections of angle bisectors of Δ C₁C₂C₃

\(\therefore\) P is incentre of the Δ C₁C₂C₃.

Answer 2

Since, the circles with radii 3, 4 and 5 touch each other externally and P is the point of intersection of tangents

=> P is incentre of ΔC₁C₂C₃.

Thus, distance of point P from the points of contact

= In radius (r) of ΔC₁C₂C₃.

Comments

Incentre hai kaise pata chala?