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

Q: Why do we need the normal line to find the centers?

The normal line is the locus of all centers of circles tangent to a given line at a specific point. Because the radius is always perpendicular to the tangent at the point of contact, the centers must lie on this perpendicular line, which we call the normal.

Q: Why is the slope of the normal $\frac{3}{4}$?

The tangent equation $4x + 3y = 10$ can be rewritten as $y = -\frac{4}{3}x + \frac{10}{3}$, so its slope $m_t$ is $-\frac{4}{3}$. Since the normal is perpendicular to the tangent, its slope $m_n$ is the negative reciprocal: $m_n = -\frac{1}{m_t} = -\frac{1}{-4/3} = \frac{3}{4}$.

Q: How do we get $\sin \theta$ and $\cos \theta$ from the slope?

Since the slope of the normal is $\tan \theta = \frac{3}{4}$, we can visualize a right-angled triangle where the opposite side is $3$ and the adjacent side is $4$. By the Pythagorean theorem, the hypotenuse is $\sqrt{3^2 + 4^2} = 5$. Therefore, $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5}$ and $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{5}$.

Question

Two circles each of radius 5 units touch each other at the point $(1, 2)$. If the equation of their common tangent is $4x + 3y = 10$, and $C_1(\alpha, \beta)$ and $C_2(\gamma, \delta)$, $C_1 
eq C_2$ are their centres, then $|(\alpha + \beta)(\gamma + \delta)|$ is equal to .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the JEE journey. Today, we stand before the elegance of two circles kissing at a single point.

Imagine the point $P(1, 2)$, where two circles of radius $r=5$ meet. They share a common tangent, $4x + 3y = 10$, which acts as a boundary at their point of contact.

To find the centers $C_1$ and $C_2$, we must look beyond the circles themselves and focus on the line that connects them: the normal line.

### The Normal Line: The Spine of the Problem

In geometry, the radius of a circle is always perpendicular to the tangent at the point of contact. This means the centers of our two circles must lie on a line that passes through $P(1, 2)$ and is perpendicular to the common tangent.

First, let us find the slope of the tangent. By rearranging $4x + 3y = 10$ into the slope-intercept form $y = mx + c$, we get:

$$y = -\frac{4}{3}x + \frac{10}{3}$$

The slope $m_t$ is $-\frac{4}{3}$. Since the normal is perpendicular to this tangent, its slope $m_n$ is the negative reciprocal:

$$m_n = \frac{3}{4}$$

This is the path our centers must follow.

### The Parametric Voyage

Now, how do we find the coordinates of the centers? We know they are at a distance $r=5$ from the point $P(1, 2)$ along the normal line.

This is the perfect moment to use the parametric form of a line. If the slope of the line is $	an 	heta = \frac{3}{4}$, we can construct a right triangle with opposite side $3$ and adjacent side $4$, making the hypotenuse $5$.

Thus, $\sin 	heta = \frac{3}{5}$ and $\cos 	heta = \frac{4}{5}$. The coordinates of the centers are given by $(x_1 \pm r \cos 	heta, y_1 \pm r \sin 	heta)$.

Substituting our values, we get:

$$x = 1 \pm 5\left(\frac{4}{5}ight)$$

$$y = 2 \pm 5\left(\frac{3}{5}ight)$$

### The Final Calculation

Let us calculate the first center, $C_1$, by taking the positive sign: $x = 1 + 4 = 5$ and $y = 2 + 3 = 5$. So, $C_1 = (5, 5)$.

Now, for the second center, $C_2$, we take the negative sign: $x = 1 - 4 = -3$ and $y = 2 - 3 = -1$. So, $C_2 = (-3, -1)$.

We have found our parameters: $\alpha=5, \beta=5, \gamma=-3, \delta=-1$. The problem asks for $|(\alpha + \beta)(\gamma + \delta)|$.

Calculating the sums, we find $(\alpha + \beta) = 10$ and $(\gamma + \delta) = -4$. The product is:

$$(10)(-4) = -40$$

Taking the absolute value, we arrive at our final, rock-solid answer: **40**.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Slope of the Common Tangent

The Normal Line

Trigonometric Ratios of the Normal

Parametric Form of Centers

Setting up Center $C_{1}$

Calculating Center $C_{1}$

Setting up Center $C_{2}$

Calculating Center $C_{2}$

Final Expression Setup

Evaluating the Sums

The Final Result

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Normal Line

The Parametric Voyage

The Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Slope of the Common Tangent

The Normal Line

Trigonometric Ratios of the Normal

Parametric Form of Centers

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

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

Setting up Center .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

Calculating Center .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

Final Expression Setup

Evaluating the Sums

The Final Result

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Normal Line

The Parametric Voyage

The Final Calculation

Similar Questions

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Setting up Center $C_{1}$

Calculating Center $C_{1}$

Setting up Center $C_{2}$

Calculating Center $C_{2}$

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