Two circles, each of radius 5 units, touch each other at (1, 2). If the equation of their common tan

Q: Wait, why do we use the normal line to find the centers? Can't we just use the distance formula from the tangent?

That is a brilliant intuition! You could use the distance formula, but it would lead to a system of equations that is algebraically heavier. By using the normal line, we leverage the geometric property that the radius is always perpendicular to the tangent at the point of contact. It turns a complex algebraic problem into a simple vector-like displacement.

Q: Why are there two circles? Does the problem imply they are on opposite sides of the tangent?

Exactly! When two circles touch at a point, they can either lie on opposite sides of the common tangent or on the same side (though if they are on the same side, they would be identical). The $\pm$ in our parametric equation naturally accounts for the two possible directions along the normal line, giving us both centers.

Q: I got the centers right, but my final circle equations look different. Did I expand them wrong?

Don't panic! Expanding $(x-h)^2 + (y-k)^2 = r^2$ is where most students make sign errors. Remember that $(x - (-3))^2$ becomes $(x+3)^2$. Take a deep breath, re-check your signs, and ensure you moved the $r^2$ term to the correct side of the equation.

Q: Is there a way to solve this using the family of circles equation?

Yes, you could use the concept of a circle touching a line at a point, but that is often more abstract. The parametric approach is the 'gold standard' for JEE because it is visual, direct, and minimizes the chance of missing a solution.

Q: Does the radius being 5 change how we calculate the normal line?

Not at all! The normal line is defined solely by the point of contact and the tangent's slope. The radius simply tells us 'how far' to walk along that normal line to find the centers. The geometry of the line is independent of the size of the circles.

Question

Two circles, each of radius 5 units, touch each other at $(1, 2)$. If the equation of their common tangent is $4x + 3y = 10$, find the equation of the circles.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing at the point $(1, 2)$ on a coordinate plane. You are looking at a line, $4x + 3y = 10$, which acts as a bridge—a common tangent—between two perfect circles, each with a radius of $r = 5$.

To find these circles, we follow the path of the normal. In the world of circles, the radius is the most loyal companion to the tangent. At the point of contact, the radius is always perpendicular to the tangent.

### The Normal: Our Compass

First, let's look at our tangent: $4x + 3y = 10$. If we rewrite this in the slope-intercept form, $y = -\frac{4}{3}x + \frac{10}{3}$, we see the slope $m_1 = -\frac{4}{3}$.

Since our normal line is perpendicular to this, its slope $m_2$ must satisfy the condition $m_1 \cdot m_2 = -1$. Thus, our normal line has a slope of $m_2 = \frac{3}{4}$.

This slope is our compass; it tells us exactly which direction to travel from our point of contact to find the hidden centers of our circles.

### The Parametric Leap

Now, we stand at $(1, 2)$ and need to move a distance of $r = 5$ along a line with slope $m = \frac{3}{4}$. We use the elegance of parametric coordinates to find the centers.

If the slope is $\frac{3}{4}$, we visualize a right triangle with a base of $4$ and a height of $3$. The hypotenuse is $\sqrt{3^2 + 4^2} = 5$. This gives us the trigonometric components:

$$\cos 	heta = \frac{4}{5}, \quad \sin 	heta = \frac{3}{5}$$

Our centers $C(h, k)$ are located at $(x_0 \pm r \cos 	heta, y_0 \pm r \sin 	heta)$. Substituting our values:

$$C = \left( 1 \pm 5 \cdot \frac{4}{5}, 2 \pm 5 \cdot \frac{3}{5} ight)$$

This simplifies beautifully to:

$$C = (1 \pm 4, 2 \pm 3)$$

### The Two Realities

By taking the positive sign, we find our first center: $C_1(5, 5)$. By taking the negative sign, we find our second center: $C_2(-3, -1)$.

We now apply the standard circle equation $(x - h)^2 + (y - k)^2 = r^2$.

For $C_1(5, 5)$, we have:

$$(x - 5)^2 + (y - 5)^2 = 25$$

Expanding this, we arrive at:

$$x^2 + y^2 - 10x - 10y + 25 = 0$$

For $C_2(-3, -1)$, we have:

$$(x + 3)^2 + (y + 1)^2 = 25$$

Expanding this, we arrive at:

$$x^2 + y^2 + 6x + 2y - 15 = 0$$

### Reflection

We started with a single point and a line, and through the power of perpendicularity and parametric motion, we reconstructed two distinct circles. This is the essence of **JEE mathematics**: taking a complex, abstract constraint and breaking it down into a sequence of logical, geometric steps.

You have successfully mapped the architecture of the plane. Keep this clarity, and the next problem will be just as conquerable.

Animated Solution for Mathematics - Circles: Two circles, each of radius 5 units, touch each other at $(1, 2)$ . If the equation of their common tangent is $4 x + 3 y = 10$ , find the equation of the circles.

Visualized Solution

Visualizing the Setup

The Normal Line Property

Slope of the Tangent

Slope of the Normal

Parametric Form of a Line

Finding $cos θ$ and $sin θ$

Substituting into Parametric Form

Calculating the Centers

Equation of Circle $1$

Expanding Circle $1$

Equation of Circle $2$

Expanding Circle $2$

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Normal

The Parametric Leap

The Two Realities

Reflection

Similar Questions

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Comprehension Passage

A possible equation of $L$ is

A common tangent of the two circles is

Let the tangents at the points $A (4, - 11)$ and $B (8, - 5)$ on the circle $x^{2} + y^{2} - 3 x + 10 y - 15 = 0$ , intersect at the point $C$ . Then the radius of the circle, whose centre is $C$ and the line joining $A$ and $B$ is its tangent, is equal to