A circle is given by x² + (y - 1)² = 1, another circle C touches it externally and also the x-axis,

Q: Wait, why is the radius $|k|$ and not just $k$? Can't a radius just be positive?

That is a fantastic question! You are absolutely right that a radius must be a positive physical length. However, the coordinate $k$ can be negative if the circle is below the x-axis. By using $|k|$, we ensure the radius remains positive regardless of where the circle sits. It is the mathematical way of saying 'distance from the x-axis'.

Q: Why do we use the sum of radii for external touch? What if they touch internally?

Great intuition! For external tangency, the circles sit side-by-side, so the distance between centers is the sum of radii ($R_1 + R_2$). If they touched internally, the distance between centers would be the difference of their radii ($|R_1 - R_2|$). The problem explicitly states 'externally', so we stick to the sum.

Q: Why do we have to split the problem into cases for $k > 0$ and $k \leq 0$?

Because of the modulus function! The expression $|k|$ behaves differently depending on the sign of $k$. If we don't split the cases, we miss half the locus. In JEE, ignoring the domain of a function is the fastest way to lose marks. Always respect the modulus!

Question

A circle is given by $x^2 + (y - 1)^2 = 1$, another circle $C$ touches it externally and also the x-axis, then the locus of its centre is

Options:

(A) $\{(x, y) : x^2 = 4y\} \cup \{(x, y) : y \leq 0\}$

(B) $\{(x, y) : x^2 + (y - 1)^2 = 4\} \cup \{(x, y) : y \leq 0\}$

(C) $\{(x, y) : x^2 = y\} \cup \{(x, y) : y \leq 0\}$

(D) $\{(x, y) : x^2 = 4y\} \cup \{(0, y) : y \leq 0\}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The fixed circle is defined by the equation $x^2 + (y - 1)^2 = 1$. By comparing this to the standard form $(x-a)^2 + (y-b)^2 = r^2$, we identify its center at $O_1(0, 1)$ and its radius $R_1 = 1$.

This circle is perfectly balanced on the $x$-axis, tangent to the origin. We now introduce a moving circle $C$ with center $P(h, k)$ and radius $r$, which must touch $C_1$ externally and also touch the $x$-axis.

### The Constraint of the Axis

If circle $C$ touches the $x$-axis, its center $(h, k)$ must be at a perpendicular distance from the $x$-axis equal to its radius $r$. Since the distance from $(h, k)$ to the $x$-axis is $|k|$, we establish the relationship:

$$r = |k|$$

This is a crucial realization. We use the modulus $|k|$ because the circle could potentially exist below the $x$-axis, where $k$ is negative.

### The Bridge of Tangency

When two circles touch externally, the distance between their centers is exactly the sum of their radii. Given the fixed center $O_1(0, 1)$ and moving center $P(h, k)$, the distance $O_1P$ is:

$$\sqrt{(h - 0)^2 + (k - 1)^2}$$

The sum of the radii is $R_1 + r = 1 + |k|$. Equating these, we obtain the fundamental equation:

$$\sqrt{h^2 + (k - 1)^2} = 1 + |k|$$

Squaring both sides to eliminate the radical yields:

$$h^2 + (k - 1)^2 = (1 + |k|)^2$$

Expanding both sides, we get:

$$h^2 + k^2 - 2k + 1 = 1 + 2|k| + |k|^2$$

Since $|k|^2 = k^2$, the $k^2$ terms and the constants cancel out, leaving us with:

$$h^2 - 2k = 2|k| \quad \Rightarrow \quad h^2 = 2k + 2|k|$$

### Resolving the Modulus

To find the final locus, we must analyze the equation $h^2 = 2k + 2|k|$ based on the sign of $k$.

**Case 1: $k > 0$**
In this scenario, $|k| = k$. The equation becomes:

$$h^2 = 2k + 2k = 4k$$

Replacing $(h, k)$ with $(x, y)$, we obtain the parabola:

$$x^2 = 4y$$

**Case 2: $k \leq 0$**
In this scenario, $|k| = -k$. The equation becomes:

$$h^2 = 2k - 2k = 0 \quad \Rightarrow \quad h = 0$$

This represents the $y$-axis for the region where $y \leq 0$.

### Final Result

The locus of the center $P(h, k)$ is the union of the parabola **$x^2 = 4y$** (for $y > 0$) and the ray **$x = 0$** (for $y \leq 0$).

Animated Solution for Mathematics - Circles: A circle is given by $x^{2} + (y - 1)^{2} = 1$ , another circle $C$ touches it externally and also the x-axis, then the locus of its centre is

Select Answer:

Visualized Solution

The Fixed Circle $C_{1}$

Center and Radius of $C_{1}$

The Moving Circle $C$

Radius of the Moving Circle

Condition for External Touch

Setting Up the Equation

Squaring Both Sides

Expanding the Terms

Simplifying the Equation

Handling the Modulus (Case Analysis)

Case 1: $k > 0$

Case 2: $k \leq 0$

The Final Locus

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Constraint of the Axis

The Bridge of Tangency

Resolving the Modulus

Final Result

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Animated Solution for Mathematics - Circles: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A circle is given by x2+(y−1)2=1, another circle C touches it externally and also the x-axis, then the locus of its centre is