Circles with radii 3, 4 and 5 touch each other externally. If P is the point of intersection of tang

Q: Wait, why is the radical center the same as the incenter of the triangle of centers?

This is a beautiful geometric coincidence! The radical center is the point where the power of the point is equal with respect to all three circles. For mutually touching circles, the common tangents at the points of contact are the radical axes. The intersection of these radical axes (the radical center) is equidistant from the points of contact. In the triangle formed by the centers, the distances from the incenter to the sides are the inradius, and these points of contact on the sides of the triangle are exactly the points where the circles touch. Thus, the radical center coincides with the incenter.

Q: Can we use coordinate geometry to solve this instead of the incenter property?

Absolutely! You could place the centers on a coordinate plane, find the equations of the circles, and solve for the radical center by setting the power of the point equal for all three. However, that approach involves solving a system of linear equations and is significantly more tedious. The geometric approach using the incenter property is the 'JEE way'—it rewards your ability to see the underlying structure of the problem.

Q: Why is the distance from the radical center to the contact point equal to the inradius?

Think about the definition of the inradius. It is the perpendicular distance from the incenter to the sides of the triangle. In our configuration, the sides of the triangle are formed by the lines connecting the centers. The points of contact lie exactly on these lines. Therefore, the distance from the incenter to these points of contact is, by definition, the perpendicular distance to the sides, which is the inradius.

Question

Circles 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 points of contact.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing in a vast, empty plane. Before you, three circles of different sizes—radii $r_1=3$, $r_2=4$, and $r_3=5$—are placed so that they kiss each other externally.

In the world of JEE Advanced geometry, this is not just a picture; it is a puzzle waiting to be unlocked. We are tasked with finding the distance from the point of intersection of the common tangents at their contact points to the points of contact themselves.

### The Triangle of Centers

To understand the relationship between these circles, we must look at their hearts—their centers. Let us label them $C_1, C_2,$ and $C_3$.

When two circles touch externally, the distance between their centers is simply the sum of their radii. By connecting these centers, we form a triangle, $	riangle C_1C_2C_3$. The side lengths of this triangle are determined by the sum of the radii of the circles they connect:

- Side $c = C_1C_2 = r_1 + r_2 = 3 + 4 = 7$
- Side $a = C_2C_3 = r_2 + r_3 = 4 + 5 = 9$
- Side $b = C_3C_1 = r_3 + r_1 = 5 + 3 = 8$

We have transformed a problem about circles into a problem about a triangle with sides $7, 8,$ and $9$. This is the power of geometric reduction.

### The Radical Center

Now, consider the tangents drawn at the points where these circles touch. These lines are the radical axes of the circles. They meet at a single, unique point $P$, known as the **Radical Center**.

A fundamental property of the radical center is that the lengths of the tangents drawn from it to any of the circles are equal. This means the distance from $P$ to each point of contact is identical.

For three mutually touching circles, the radical center $P$ is exactly the incenter of the triangle formed by their centers. Consequently, the distance we seek is the inradius $r$ of $	riangle C_1C_2C_3$.

### The Final Calculation

With the realization that we are simply looking for the inradius, the path forward is clear. First, we calculate the semi-perimeter $s$ of our triangle:

$$s = \frac{a + b + c}{2} = \frac{9 + 8 + 7}{2} = \frac{24}{2} = 12$$

Now, we employ the elegant formula for the inradius of a triangle:

$$r = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$$

Substituting our values into this expression, we get:

$$r = \sqrt{\frac{(12-9)(12-8)(12-7)}{12}}$$

Let us simplify this carefully. The numerator becomes $3 \cdot 4 \cdot 5$, which is $60$. Dividing by $12$, we find:

$$r = \sqrt{\frac{60}{12}} = \sqrt{5}$$

The final distance from the radical center to the points of contact is **$\sqrt{5}$**.

### Conclusion

What started as a complex problem involving three circles and their tangents has been distilled into the properties of a triangle. This is the essence of JEE Advanced mathematics: finding the simple, elegant truth hidden beneath a layer of complexity.

Never be intimidated by the initial appearance of a problem. Look for the underlying geometry, trust your toolkit, and the solution will reveal itself.

Animated Solution for Mathematics - Circles: Circles 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 points of contact.

Enter Numerical Value:

Visualized Solution

Visualizing the Configuration

Identifying Point $P$

The Radical Center Property

Forming the Triangle of Centers

The Incenter Connection

Calculating Semi-perimeter $s$

Applying the Inradius Formula

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Triangle of Centers

The Radical Center

The Final Calculation

Conclusion

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