Let C₁ be the circle of radius 1 with center at the origin. Let C₂ be the circle of radius r with ce

Q: Wait, why is the line joining the midpoints the Radical Axis? I thought the Radical Axis was just for intersecting circles.

That is a fantastic question! The Radical Axis is defined for any two circles, whether they intersect or not. The key insight here is that for any point on a common tangent, the length of the tangent to $C_1$ equals the length of the tangent to $C_2$. Since the midpoint of a common tangent is equidistant from the points of contact, it must have equal power with respect to both circles. Therefore, the line connecting these midpoints is indeed the Radical Axis.

Q: Why can't we just use the distance formula between the centers and the radii to find the tangent lengths?

You absolutely could, but it would be incredibly tedious! You would have to find the coordinates of the points of contact $P, Q, S, T$, then find the midpoints, and then find the line equation. The Radical Axis method is the elegant 'JEE way'—it bypasses all that heavy coordinate calculation and gives you the line equation in one step.

Q: Why did we reject the negative root when taking the square root of 4 in the final step?

Great catch! We had the equation $\left( \frac{14 + r^2}{8} \right)^2 = 4$. Taking the square root gives $\frac{14 + r^2}{8} = \pm 2$. However, we are given that $r > 1$, so $r^2 > 1$. This means $14 + r^2$ must be greater than 15. Thus, the fraction $\frac{14 + r^2}{8}$ must be positive. Therefore, we only consider the positive root, $2$.

Question

Let $C_1$ be the circle of radius 1 with center at the origin. Let $C_2$ be the circle of radius $r$ with center at the point $A = (4,1)$, where $1 < r < 3$. Two distinct common tangents $PQ$ and $ST$ of $C_1$ and $C_2$ are drawn. The tangent $PQ$ touches $C_1$ at $P$ and $C_2$ at $Q$. The tangent $ST$ touches $C_1$ at $S$ and $C_2$ at $T$. Mid points of the line segments $PQ$ and $ST$ are joined to form a line which meets the x-axis at a point $B$. If $AB = \sqrt{5}$, then the value of $r^2$ is

Sigma Percentile · Accepted Answer

### The Geometry of Tangents: A Radical Insight

Welcome, fellow explorer of the coordinate plane! Today, we are going to unravel a problem that, at first glance, seems to demand a mountain of tedious coordinate geometry.

We have two circles, $C_1$ and $C_2$, and we are drawing common tangents. The problem asks us to find the value of $r^2$ given a specific distance condition involving the midpoints of these tangents.

If you try to find the coordinates of the points of contact $P, Q, S,$ and $T$ directly, you will find yourself drowning in a sea of algebra. But fear not! There is a beautiful, hidden geometric shortcut that will turn this nightmare into a dream.

### The Setup: Defining Our Circles

Let us start by defining our territory. We have $C_1$, a circle of radius $1$ centered at the origin. Its equation is straightforward:
$$x^2 + y^2 = 1$$

Then we have $C_2$, centered at $A = (4, 1)$ with an unknown radius $r$. Its equation is $(x-4)^2 + (y-1)^2 = r^2$.

Expanding this, we get $x^2 - 8x + 16 + y^2 - 2y + 1 = r^2$, or:
$$x^2 + y^2 - 8x - 2y + 17 - r^2 = 0$$
This is our foundation.

### The 'Aha!' Moment: The Radical Axis

Now, consider the common tangents $PQ$ and $ST$. The problem mentions the midpoints of these segments. Let $M_1$ be the midpoint of $PQ$ and $M_2$ be the midpoint of $ST$.

Here is the secret: the length of the tangent from any point on the Radical Axis to both circles is equal. For the midpoint of a common tangent, the distance to the point of contact on $C_1$ is equal to the distance to the point of contact on $C_2$.

This means the midpoint lies on the Radical Axis! Since both midpoints $M_1$ and $M_2$ lie on this line, the line joining them *is* the Radical Axis.

### The Algebra of Elegance

To find the equation of the Radical Axis, we simply subtract the two circle equations: $S_1 - S_2 = 0$. Let $S_1 = x^2 + y^2 - 1 = 0$ and $S_2 = x^2 + y^2 - 8x - 2y + 17 - r^2 = 0$.

Subtracting them, the quadratic terms $x^2$ and $y^2$ vanish, leaving us with:
$$(x^2 + y^2 - 1) - (x^2 + y^2 - 8x - 2y + 17 - r^2) = 0$$

This simplifies beautifully to:
$$8x + 2y - 18 + r^2 = 0$$
This is the equation of the line joining the midpoints.

### The Intersection and the Final Stretch

We are told this line meets the x-axis at point $B$. On the x-axis, $y = 0$.

Substituting this into our line equation, we get $8x - 18 + r^2 = 0$, which gives:
$$x = \frac{18 - r^2}{8}$$
So, $B = \left( \frac{18 - r^2}{8}, 0 \right)$.

We are given $AB = \sqrt{5}$, so $AB^2 = 5$. Using the distance formula with $A = (4, 1)$ and $B = \left( \frac{18 - r^2}{8}, 0 \right)$, we have:
$$\left( 4 - \frac{18 - r^2}{8} \right)^2 + (1 - 0)^2 = 5$$

Simplifying the term inside the square, we get:
$$\left( \frac{32 - 18 + r^2}{8} \right)^2 + 1 = 5$$
$$\left( \frac{14 + r^2}{8} \right)^2 = 4$$

Taking the square root, we get $\frac{14 + r^2}{8} = 2$. Solving for $r^2$, we find $14 + r^2 = 16$, so:
**$r^2 = 2$**

### Conclusion

And there it is! By recognizing the Radical Axis, we bypassed the grueling task of finding tangent points and midpoints.

Geometry is not just about calculation; it is about seeing the underlying structure. Keep practicing, keep visualizing, and you will find that even the most intimidating problems have a simple, elegant heart.

Enter Numerical Value:

Visualized Solution

Circles $C_{1}$ and $C_{2}$

Common Tangents

The Radical Axis Concept

Equation of Radical Axis

Simplifying Radical Axis

Intersection with x-axis

Coordinates of Point B

Distance Condition

Applying Distance Formula

Simplifying the Equation

Solving for $r^{2}$

Final Answer

The Sigma Insight: Director Circle and Family of Circles

The Geometry of Tangents

The Setup

The 'Aha!' Moment

The Algebra of Elegance

The Intersection and the Final Stretch

Conclusion

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