The centres of two circles C₁ and C₂ each of unit radius are at a distance of 6 units from each othe

Q: Wait, why do we assume the center of circle $C$ is on the y-axis?

Great question! Because circles $C_1$ and $C_2$ are placed symmetrically about the y-axis (at $x = -3$ and $x = 3$), any circle $C$ that touches both of them externally must also be symmetric with respect to the y-axis. If the center were not on the y-axis, the setup would lose its symmetry, and the circle would not touch both $C_1$ and $C_2$ in the same way. Symmetry is a powerful tool in coordinate geometry!

Q: Why does the tangent line pass through the origin?

The problem explicitly states that the common tangent passes through point $P$, which is the midpoint of the line segment joining the centers of $C_1$ and $C_2$. Since we set the centers at $(-3,0)$ and $(3,0)$, the midpoint $P$ is indeed $(0,0)$. Thus, any line passing through $P$ can be represented as $y = mx$.

Q: Is there a way to solve this without using coordinate geometry?

While you could potentially use pure geometry or inversion, coordinate geometry is the most robust and systematic approach for this specific problem. It turns a complex geometric visualization into a clear algebraic system of equations. Stick with the coordinate method; it is less prone to errors and highly reliable for JEE Advanced.

Question

The centres of two circles $C_1$ and $C_2$ each of unit radius are at a distance of 6 units from each other. Let $P$ be the mid point of the line segement joining the centres of $C_1$ and $C_2$ and $C$ be a circle touching circles $C_1$ and $C_2$ externally. If a common tangent to $C_1$ and $C$ passing through $P$ is also a common tangent to $C_2$ and $C$, then the radius of the circle $C$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To solve this problem, we establish a coordinate system by placing the midpoint $P$ of the segment connecting the centers of $C_1$ and $C_2$ at the origin $(0,0)$. Given the distance between the centers is $6$, we position the centers of $C_1$ and $C_2$ at $(-3,0)$ and $(3,0)$ respectively.

Due to the symmetry of the configuration across the $y$-axis, the center of the third circle $C$ must lie on the $y$-axis. Let the center of circle $C$ be $(0,k)$ and its radius be $r$.

### The Tangent's Secret

Consider a common tangent line passing through the origin $P(0,0)$. Its equation is $y = mx$, or in standard form, $mx - y = 0$.

For circle $C_1$ (center $(-3,0)$, radius $1$), the perpendicular distance from the center to the line must equal the radius:

$$\frac{|m(-3) - 0|}{\sqrt{m^2 + 1}} = 1$$

Squaring both sides yields $9m^2 = m^2 + 1$, which simplifies to:

$$m^2 = \frac{1}{8}$$

Now, we apply the same condition to circle $C$ (center $(0,k)$, radius $r$):

$$\frac{|m(0) - k|}{\sqrt{m^2 + 1}} = r$$

Squaring this expression gives $k^2 = r^2(m^2 + 1)$. Substituting $m^2 = \frac{1}{8}$ into this equation, we obtain:

$$k^2 = r^2\left(\frac{1}{8} + 1\right) = \frac{9r^2}{8}$$

### The Algebraic Finale

Since circle $C$ touches circle $C_1$ externally, the distance between their centers must equal the sum of their radii, $r + 1$. Using the distance formula between $(0,k)$ and $(-3,0)$:

$$\sqrt{(0 - (-3))^2 + (k - 0)^2} = r + 1$$

Squaring both sides results in $9 + k^2 = (r + 1)^2$. Substituting our previous expression for $k^2$ into this equation gives:

$$9 + \frac{9r^2}{8} = (r + 1)^2$$

Multiplying the entire equation by $8$ to clear the fraction, we get:

$$72 + 9r^2 = 8(r^2 + 2r + 1)$$

Expanding and rearranging the terms leads to:

$$72 + 9r^2 = 8r^2 + 16r + 8$$

$$r^2 - 16r + 64 = 0$$

This is a perfect square trinomial, which factors as $(r - 8)^2 = 0$. Therefore, the radius of the circle $C$ is **$r = 8$**.

Enter Numerical Value:

Visualized Solution

Coordinate System Setup

Symmetry of Circle $C$

External Touch Condition

Forming Equation 1

The Common Tangent

Distance from $C_{1}$ to Tangent

Solving for $m^{2}$

Distance from $C$ to Tangent

Forming Equation 2

Substitution and Algebra

Solving the Quadratic

Final Result

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent's Secret

The Algebraic Finale

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