Three Circles with centers C1, C2, C3 are tangents to each other. PC is a common tangent to all three circles which touch circles with center C1 and C2 at A, B respectively. Find the relation between PA.PB and PC?

To solve the problem involving three circles with centers C1, C2, and C3 that are tangent to each other, and a common tangent PC that touches the circles at points A and B, we can utilize some geometric properties related to tangents and circles. Let's break down the relationship between the segments PA, PB, and PC.

Understanding the Setup

We have three circles, each tangent to one another. The tangent line PC touches circle C1 at point A and circle C2 at point B. The tangents from an external point to a circle have a special property: they are equal in length from that point to the points of tangency. This will be crucial for our calculations.

Identifying Key Relationships

Let’s denote the radius of circle C1 as r1, the radius of circle C2 as r2, and the distance between the centers C1 and C2 as d. When you draw tangents from point P to the circles at points A and B, we can express the lengths of these tangents using the following properties:

• Length of Tangent from Point P to Circle C1: PA = √(PC^2 - r1^2)
• Length of Tangent from Point P to Circle C2: PB = √(PC^2 - r2^2)

Establishing the Relation

Next, we apply the idea that the product of the lengths of the tangents from a point outside the circles (in this case, point P) to the points of tangency A and B can be expressed as:

PA × PB = PC^2 - (r1 × r2)

Deriving the Product PA × PB

Since we know PA and PB are the lengths of the tangents, we can substitute the expressions we derived earlier:

PA × PB = √(PC^2 - r1^2) × √(PC^2 - r2^2)

When we simplify this, we get:

PA × PB = √[(PC^2 - r1^2)(PC^2 - r2^2)]

Final Relationship

From the derived expressions and the properties of tangents, we arrive at the final relationship:

PA × PB = PC^2 - r1 × r2

This relationship illustrates how the segments PA and PB interact with the tangent segment PC and the radii of the circles. It’s a beautiful example of how geometry and algebra intertwine to explain the relationships in a problem involving tangents to circles.

In summary, the relationship between PA, PB, and PC can be succinctly expressed as:

PA × PB = PC^2 - r1 × r2

This equation highlights the harmony in geometric relationships, and recognizing these connections can deepen your understanding of circle geometry and tangents.

Approved