Let S1 : x² + y² = 9 and S2 : (x - 2)² + y² = 1. Then the locus of center of a variable circle S whi

Q: Wait, why do we add the distances $PA$ and $PB$? Shouldn't we subtract them?

Great observation! We add them because the variable circle $S$ is 'trapped' between the two conditions. Since $S$ touches $S_1$ internally, the distance $PA$ is $r_1 - r$. Since it touches $S_2$ externally, $PB$ is $r_2 + r$. When you add these, the variable radius $r$ cancels out perfectly, leaving you with a constant sum, which is the hallmark of an ellipse.

Q: How do I know for sure that the locus is an ellipse and not a hyperbola?

The definition of an ellipse is the locus of points where the sum of distances to two fixed points is constant ($PA + PB = 2a$). If the difference of distances were constant ($|PA - PB| = 2a$), you would be looking at a hyperbola. Here, the geometry forces a sum, so it must be an ellipse!

Q: Do I really need to calculate the eccentricity $e$ to find the equation?

You don't strictly need $e$ if you already know $a$ and the distance between the foci ($2ae$). Once you have $a=2$ and the distance between foci $2ae = 2$, you find $e=1/2$. Then, $b^2 = a^2(1-e^2)$ is just the standard tool to find the minor axis. It's a very reliable path to the final equation.

Q: Why is the center of the ellipse $(1,0)$ and not $(0,0)$?

The center of the ellipse is always the midpoint of the line segment connecting the two foci. Since your foci are at $A(0,0)$ and $B(2,0)$, the midpoint is clearly $(\frac{0+2}{2}, \frac{0+0}{2}) = (1,0)$. Always look at where your fixed points are located!

Q: Is there any scenario where this locus could be a circle?

Only if the two foci were at the same point (i.e., the distance between them was zero). In this problem, the foci are distinct points at $(0,0)$ and $(2,0)$, so the locus must be an ellipse. If the foci were coincident, the ellipse would degenerate into a circle.

Question

Let $S_1 : x^2 + y^2 = 9$ and $S_2 : (x - 2)^2 + y^2 = 1$. Then the locus of center of a variable circle $S$ which touches $S_1$ internally and $S_2$ externally always passes through the points:

Options:

(A) $(0, \pm\sqrt{3})$

(B) $(\frac{1}{2}, \pm\frac{\sqrt{5}}{2})$

(C) $(2, \pm\frac{3}{2})$

(D) $(1, \pm 2)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, empty plane. You have two fixed anchors: a large circle $S_1$ centered at the origin $(0,0)$ with a radius of $3$, and a smaller circle $S_2$ centered at $(2,0)$ with a radius of $1$.

Now, imagine a third, smaller circle $S$ that is constantly moving. This circle is a traveler; it touches the inner boundary of the large circle $S_1$ and simultaneously kisses the outer boundary of the small circle $S_2$.

As this circle $S$ moves, its center $P(x,y)$ traces a path. Today, we are going to uncover the hidden geometry behind this motion.

### The Geometry of Constraints

Let the center of our variable circle $S$ be $P(x,y)$ and its radius be $r$. When a circle touches another internally, the distance between their centers is the difference of their radii.

Thus, for $S$ and $S_1$, we have the distance $PA = |r_1 - r| = 3 - r$. Conversely, when a circle touches another externally, the distance between their centers is the sum of their radii.

For $S$ and $S_2$, we have $PB = r_2 + r = 1 + r$.

### The Master Equation

Look at these two equations. We have a variable $r$ that we don't really care about—it's just a parameter of the circle's size. To find the locus of $P$, we need to eliminate $r$.

If we add these two equations together, something magical happens:

$$PA + PB = (3 - r) + (1 + r) = 4$$

The $r$ terms vanish, leaving us with a beautiful, constant sum: $PA + PB = 4$.

### The Birth of the Ellipse

In the language of coordinate geometry, the locus of a point such that the sum of its distances from two fixed points (the foci) is constant is the definition of an ellipse. Our fixed points are $A(0,0)$ and $B(2,0)$.

The constant sum is $2a = 4$, which tells us that the semi-major axis $a$ is $2$. Since the foci are at $(0,0)$ and $(2,0)$, the center of our ellipse must be the midpoint of the segment $AB$, which is $(1,0)$.

The distance between the foci is $2ae = 2$. With $a=2$, we find the eccentricity $e = \frac{1}{2}$.

Now, we need the minor axis $b$. Using the fundamental relation for an ellipse, $b^2 = a^2(1 - e^2)$, we calculate:

$$b^2 = 4\left(1 - \frac{1}{4}\right) = 4\left(\frac{3}{4}\right) = 3$$

### The Final Equation

With the center $(h,k) = (1,0)$, $a^2 = 4$, and $b^2 = 3$, we can write the standard equation of our locus:

$$\frac{(x-1)^2}{4} + \frac{y^2}{3} = 1$$

This equation represents the entire path of the center $P$. To verify which points lie on this path, we simply test the options provided.

Taking the point $(2, \pm \frac{3}{2})$, we substitute these coordinates into our equation:

$$\frac{(2-1)^2}{4} + \frac{(\pm \frac{3}{2})^2}{3} = \frac{1}{4} + \frac{9/4}{3} = \frac{1}{4} + \frac{3}{4} = 1$$

The equation holds true!

### Reflection

Isn't it fascinating? What started as a complex problem of moving circles resolved into the elegant, symmetric form of an ellipse.

Whenever you face a problem involving moving objects and constraints, look for the quantities that cancel out. Often, the physics or geometry is hiding a simple, beautiful truth waiting for you to uncover it.

Keep exploring, keep questioning, and most importantly, keep enjoying the process of discovery.

Animated Solution for Mathematics - Conic Sections: Let $S_{1} : x^{2} + y^{2} = 9$ and $S_{2} : (x - 2)^{2} + y^{2} = 1$ . Then the locus of center of a variable circle $S$ which touches $S_{1}$ internally and $S_{2}$ externally always passes through the points:

Select Answer:

Visualized Solution

Identify $S_{1}$ and $S_{2}$

The Variable Circle $S$

Internal Touching Condition

External Touching Condition

Eliminating the Variable Radius

Sum of Distances

Identifying the Locus

Ellipse Parameters: Center and Major Axis

Ellipse Parameters: Eccentricity

Ellipse Parameters: Minor Axis

Formulate the Locus Equation

Final Locus Equation

Verify the Passing Points

Checking the Point

Conclusion

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Geometry of Constraints

The Master Equation

The Birth of the Ellipse

The Final Equation

Reflection

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Comprehension Passage

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