Let T₁, T₂ be two tangents drawn from (-2, 0) onto the circle C : x² + y² = 1. Determine the circles

Q: Why do we assume the center of the new circle lies on the x-axis? Can't it be somewhere else?

That is a brilliant question! Because the tangents $T_1$ and $T_2$ are symmetric with respect to the x-axis, any circle that touches both of them simultaneously must also respect that symmetry. If the center were off the x-axis, the circle would be closer to one tangent than the other, violating the condition of touching both. Symmetry is your best friend in JEE geometry—use it to reduce variables!

Q: How do we know there are exactly two circles? Could there be more?

Think about the physical space available. We have the original circle at the origin and a point $P$ at $(-2, 0)$. One circle can be formed 'outside' the original circle (the larger one), and another can be 'squeezed' into the gap between the point $P$ and the original circle (the smaller one). Algebraically, this corresponds to the two different cases for the absolute value $|h| = R + 1$. There are no other regions where a circle can touch both tangents and the original circle.

Q: Why do we use the internal center of similitude for the transverse tangents?

Great observation. Transverse tangents are the ones that cross between the two circles. By definition, these tangents intersect at the internal center of similitude, which divides the line segment connecting the centers of the two circles internally in the ratio of their radii. Using this point allows us to write the equation of the tangent line in terms of a single slope parameter $M$, making the calculation much cleaner.

Question

Let $T_1, T_2$ be two tangents drawn from $(-2, 0)$ onto the circle $C : x^2 + y^2 = 1$. Determine the circles touching $C$ and $T_1, T_2$ as their pair of tangents. Further, find the equations of all possible common tangents to these circles, when taken two at a time.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are orchestrating a dance of circles. We have a unit circle, a point in space, and a set of constraints that force new circles into existence.

At the origin, we have our unit circle $C : x^2 + y^2 = 1$. Sitting at $P(-2, 0)$ is our point of origin for the tangents. We draw two tangents, $T_1$ and $T_2$, from $P$ to $C$.

Because $P$ lies on the x-axis, the entire setup is perfectly symmetric. If we draw a line through $P$ with slope $m$, its equation is $y = m(x + 2)$, or $mx - y + 2m = 0$.

### The Tangents as Our Foundation

The perpendicular distance from the center $(0, 0)$ to this line must equal the radius, which is $1$. Using the distance formula, we have:

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

This simplifies to $\frac{|2m|}{\sqrt{m^2 + 1}} = 1$. Squaring both sides, we find $4m^2 = m^2 + 1$, leading us to $3m^2 = 1$, or $m = \pm \frac{1}{\sqrt{3}}$.

Thus, our tangents are $x \mp \sqrt{3}y + 2 = 0$. We have established our boundaries.

### The Birth of New Circles

The problem asks us to find circles that touch $C$ and both $T_1$ and $T_2$. Because of the symmetry, the centers of these new circles must lie on the x-axis. Let the center be $Q(h, 0)$ and the radius be $R$.

For these circles to touch the tangents $x \mp \sqrt{3}y + 2 = 0$, the distance from $(h, 0)$ to the line must be $R$. Applying the distance formula:

$$ \frac{|h + 2|}{\sqrt{1^2 + (-\sqrt{3})^2}} = R \Rightarrow R = \frac{|h + 2|}{2} $$

These circles must touch the original circle $C$ externally. The distance between centers $(0, 0)$ and $(h, 0)$ is $|h|$. For external contact, this distance must equal the sum of the radii: $|h| = R + 1$.

By solving for the two cases—where the center is to the right of the origin ($h > 0$) and where it is squeezed between $P$ and $C$ ($-2 < h < 0$)—we find our two circles:

$$ (x - 4)^2 + y^2 = 3^2 \quad 	ext{and} \quad \left(x + \frac{4}{3}ight)^2 + y^2 = \left(\frac{1}{3}ight)^2 $$

### The Transverse Tangents

We now seek the common transverse tangents. These lines cross between the circles and must intersect at the internal center of similitude, $P'$. This point divides the segment connecting the centers $(4, 0)$ and $(-\frac{4}{3}, 0)$ in the ratio of their radii, $3 : \frac{1}{3}$, which is $9 : 1$.

Using the section formula, the x-coordinate of $P'$ is:

$$ \frac{9(-\frac{4}{3}) + 1(4)}{9 + 1} = -\frac{4}{5} $$

So, $P'$ is at $(-\frac{4}{5}, 0)$. Any line through this point has the form $y = M(x + \frac{4}{5})$, or $Mx - y + \frac{4M}{5} = 0$.

For this to be tangent to the larger circle $(x - 4)^2 + y^2 = 9$, the distance from $(4, 0)$ to this line must be $3$:

$$ \frac{|4M + \frac{4M}{5}|}{\sqrt{M^2 + 1}} = 3 \Rightarrow \frac{24|M|}{5\sqrt{M^2 + 1}} = 3 $$

This reduces to $8|M| = 5\sqrt{M^2 + 1}$. Squaring gives $64M^2 = 25(M^2 + 1)$, so $39M^2 = 25$, and $M = \pm \frac{5}{\sqrt{39}}$.

The transverse tangents are **$y = \pm \frac{5}{\sqrt{39}}(x + \frac{4}{5})$**. You have successfully navigated the geometry, the algebra, and the symmetry.

Visualized Solution

Visualizing the Setup

Equation of Tangents $T_{1}, T_{2}$

Applying the Distance Formula

Solving for Slope $m$

Defining the Required Circles

Condition 1: Touching the Tangents

Condition 2: Touching Circle $C$

Solving for the First Circle ( $h > 0$ )

Solving for the Second Circle ( $h < 0$ )

Identifying Common Tangents

Internal Center of Similitude

Equation of Transverse Tangents

Applying Distance Formula for $M$

Solving for Slope $M$

Final Equations of Transverse Tangents

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangents as Our Foundation

The Birth of New Circles

The Transverse Tangents

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Comprehension Passage

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Visualized Solution

Visualizing the Setup

Equation of Tangents .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T1​,T2​

Applying the Distance Formula

Solving for Slope .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Defining the Required Circles

Condition 1: Touching the Tangents

Condition 2: Touching Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Solving for the First Circle ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h>0)

Solving for the Second Circle ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h<0)