If a circle C, whose radius is 3, touches externally the circle, x² + y² - 4y - 4 = 0 at the point (

Q: Wait, why do we assume the center of the new circle lies on the line $y = 2$?

That is a fantastic question! When two circles touch each other externally, the point of contact and the two centers are always collinear. Since the center of the first circle $O_1$ is $(0, 2)$ and the point of contact $P$ is $(2, 2)$, the line passing through them is the horizontal line $y = 2$. Therefore, the center of the second circle must also lie on this same line $y = 2$.

Q: Does the choice of the center (left or right) change the intercept length?

Not at all! The intercept length formula $L = 2\sqrt{r^2 - k^2}$ only depends on the radius $r$ and the y-coordinate of the center $k$. Since both possible centers, $(5, 2)$ and $(-1, 2)$, have the same y-coordinate $k = 2$ and the same radius $r = 3$, the intercept length will be identical in both cases. It is a beautiful example of geometric symmetry.

Q: Can I just use the general equation of a circle to find the intercept?

You absolutely can! You could write the equation of the circle as $(x - h)^2 + (y - 2)^2 = 3^2$, set $y = 0$ to find the x-intercepts, and calculate the distance between the roots. However, the geometric shortcut $L = 2\sqrt{r^2 - k^2}$ is much faster and less prone to algebraic errors. In competitive exams like JEE, mastering these geometric shortcuts is key to saving time.

Question

If a circle C, whose radius is 3, touches externally the circle, $x^2 + y^2 - 4y - 4 = 0$ at the point $(2, 2)$, then the length of the intercept cut by this circle C, on the x-axis is equal to :

Options:

(A) $2\sqrt{3}$

(B) $3\sqrt{2}$

(C) $\sqrt{5}$

(D) $2\sqrt{5}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, watching two circles perform a delicate dance. One is fixed, anchored at a specific location, and the other is approaching it, destined to kiss it at a single, precise point.

We are given a circle $C_1$ defined by the equation $x^2 + y^2 - 4y - 4 = 0$. Our mission is to find the length of the intercept that a second circle $C$, with a radius of $3$, cuts on the x-axis after touching $C_1$ externally at the point $P(2, 2)$.

### Unmasking the First Circle

Before we can understand the new circle, we must fully comprehend the one we are given. The equation $x^2 + y^2 - 4y - 4 = 0$ can be rewritten by completing the square for the y-terms:

$$x^2 + (y^2 - 4y + 4) - 4 - 4 = 0$$

This simplifies to the standard form:

$$x^2 + (y - 2)^2 = 8$$

Comparing this to the standard form $(x - h)^2 + (y - k)^2 = r^2$, we see that the center of our first circle, $O_1$, is at $(0, 2)$, and its radius squared is $8$.

### The Secret of Collinearity

Now, consider the point of contact $P(2, 2)$. The center $O_1(0, 2)$ and the point of contact $P(2, 2)$ share the same y-coordinate, $y = 2$.

When two circles touch, their centers and the point of contact are always collinear. This means the line connecting the center of the first circle and the point of contact is the horizontal line $y = 2$.

Consequently, the center of our new circle $C$ must also lie on this line $y = 2$. We have just unlocked the location of our new center.

### Locating the Center of Circle $C$

We know the center of circle $C$ lies on the line $y = 2$. Let us call this center $O(h, 2)$. We are given that the radius of circle $C$ is $r = 3$.

Since the circles touch at $P(2, 2)$, the distance from the center $O$ to the point of contact $P$ must be exactly the radius, $3$. The distance between $(h, 2)$ and $(2, 2)$ is $|h - 2|$.

Setting this equal to $3$, we find $|h - 2| = 3$, which gives us two possible locations for the center: $h = 5$ or $h = -1$. Whether the circle is to the right or the left of the contact point, the geometry remains the same.

### The Elegant Shortcut to the Intercept

We need the length of the intercept cut by circle $C$ on the x-axis. Imagine dropping a perpendicular from the center $O(h, 2)$ to the x-axis.

This forms a right-angled triangle where the hypotenuse is the radius $r = 3$, and one leg is the vertical distance from the center to the x-axis, which is $k = 2$. The other leg, $a$, represents half the length of the intercept.

By the Pythagorean theorem, $a^2 + k^2 = r^2$, so $a = \sqrt{r^2 - k^2}$. The total length of the intercept is:

$$L = 2a = 2\sqrt{r^2 - k^2}$$

### The Final Calculation

Substituting our known values, $r = 3$ and $k = 2$, into our formula, we get:

$$L = 2\sqrt{3^2 - 2^2}$$

This simplifies to:

$$L = 2\sqrt{9 - 4} = 2\sqrt{5}$$

It is truly beautiful how the complexity of the circle's position collapses into such a simple, clean result. The final length of the intercept is **$2\sqrt{5}$**.

Animated Solution for Mathematics - Circles: If a circle C, whose radius is 3, touches externally the circle, $x^{2} + y^{2} - 4 y - 4 = 0$ at the point $(2, 2)$ , then the length of the intercept cut by this circle C, on the x-axis is equal to :

Select Answer:

Visualized Solution

Problem Setup

Center of $C_{1}$

Point of Contact $P$

Collinearity of Centers

Coordinates of Center $O$

Solving for $h$

Visualizing Circle $C$

X-Intercept Concept

Substituting Values

Calculating the Length

Final Answer

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

Unmasking the First Circle

The Secret of Collinearity

Locating the Center of Circle $C$

The Elegant Shortcut to the Intercept

The Final Calculation

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Select Answer:

Visualized Solution

Problem Setup

Center of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

Point of Contact .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Collinearity of Centers

Coordinates of Center .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } O

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

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