Let the centre of a circle C be (α, β) and its radius r

Q: Why do we have to use the modulus in the distance formula? Can't we just drop it?

The modulus is the geometric soul of the distance formula! It ensures the distance is always positive, regardless of which side of the line the center lies on. Dropping it blindly ignores the two possible regions defined by the line. In coordinate geometry, the modulus is your safety net against negative lengths.

Q: Wait, why did we reject the second case of the modulus equation? Is there a rule?

Great catch! We didn't reject it arbitrarily. The problem explicitly stated $r < 8$. When you solve the second case, you will find an $r$ value that violates this constraint. In JEE, constraints are not just suggestions—they are the boundaries of your solution space. Always check your final results against the problem's initial conditions.

Q: Why can't we just use the slope-intercept form $y = mx + c$ for the tangents?

You absolutely could! But think about the efficiency. Using the distance formula directly links the center $(\alpha, \beta)$ to the radius $r$ in one step. In the exam, time is your currency—choose the path that minimizes algebraic friction and keeps your focus on the core geometry.

Question

Let the centre of a circle $C$ be $(\alpha, \beta)$ and its radius $r < 8$. Let $3x + 4y = 24$ and $3x - 4y = 32$ be two tangents and $4x + 3y = 1$ be a normal to $C$. Then $(\alpha - \beta + r)$ is equal to

Options:

(A) 7

(B) 5

(C) 6

(D) 9

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we aren't just solving a coordinate geometry problem; we are uncovering the hidden architecture of a circle. Imagine you are standing on the Cartesian plane with a circle centered at $(\alpha, \beta)$ and a radius $r$.

We are given three lines: two tangents, $3x + 4y = 24$ and $3x - 4y = 32$, and a normal, $4x + 3y = 1$. Our mission is to find the value of $(\alpha - \beta + r)$.

### The Anchor of the Normal

Every circle has a heartbeat, and that heartbeat is its center. We are given a normal line, $4x + 3y = 1$. A normal to a circle is a line that passes through the center.

If the center $(\alpha, \beta)$ lies on this line, then the coordinates must satisfy the equation. This gives us our first solid piece of the puzzle:

$$4\alpha + 3\beta = 1$$

This linear relationship constrains the center, ensuring it is tethered to this specific line.

### The Tangency Constraint

A tangent is a line that touches the circle at exactly one point. Geometrically, the perpendicular distance from the center $(\alpha, \beta)$ to the tangent line must be exactly equal to the radius $r$. We use the point-to-line distance formula:

$$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$

For our first tangent, $3x + 4y - 24 = 0$, the distance is:

$$r = \frac{|3\alpha + 4\beta - 24|}{\sqrt{3^2 + 4^2}} = \frac{|3\alpha + 4\beta - 24|}{5}$$

For our second tangent, $3x - 4y - 32 = 0$, the distance is:

$$r = \frac{|3\alpha - 4\beta - 32|}{\sqrt{3^2 + (-4)^2}} = \frac{|3\alpha - 4\beta - 32|}{5}$$

### The Algebraic Dance

Since both expressions equal $r$, we equate them:

$$\frac{|3\alpha + 4\beta - 24|}{5} = \frac{|3\alpha - 4\beta - 32|}{5}$$

The denominators cancel out, leaving us with a modulus equation:

$$|3\alpha + 4\beta - 24| = |3\alpha - 4\beta - 32|$$

Considering the case where the expressions inside the modulus are equal:

$$3\alpha + 4\beta - 24 = 3\alpha - 4\beta - 32$$

The $3\alpha$ terms cancel out, simplifying the equation significantly:

$$4\beta + 4\beta = -32 + 24 \Rightarrow 8\beta = -8 \Rightarrow \beta = -1$$

Substituting $\beta = -1$ into our anchor equation $4\alpha + 3\beta = 1$:

$$4\alpha + 3(-1) = 1 \Rightarrow 4\alpha - 3 = 1 \Rightarrow 4\alpha = 4 \Rightarrow \alpha = 1$$

We have successfully identified the center as $(1, -1)$.

### Final Calculation

Now, we calculate the radius $r$ by substituting the center $(1, -1)$ into the distance formula:

$$r = \frac{|3(1) + 4(-1) - 24|}{5} = \frac{|3 - 4 - 24|}{5} = \frac{|-25|}{5} = 5$$

Finally, we compute the requested value:

$$\alpha - \beta + r = 1 - (-1) + 5 = 1 + 1 + 5 = 7$$

The final answer is **7**.

Animated Solution for Mathematics - Circles: Let the centre of a circle $C$ be $(α, β)$ and its radius $r < 8$ . Let $3 x + 4 y = 24$ and $3 x - 4 y = 32$ be two tangents and $4 x + 3 y = 1$ be a normal to $C$ . Then $(α - β + r)$ is equal to

Select Answer:

Visualized Solution

Visualizing the Circle and Lines

The Normal Property

Substituting the Center

Tangency Condition

Distance to Tangent 1

Simplifying Distance 1

Distance to Tangent 2

Equating the Distances

Solving the Absolute Value (Case 1)

Finding $β$

Finding $α$

Calculating Radius $r$

Checking Constraints

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Anchor of the Normal

The Tangency Constraint

The Algebraic Dance

Final Calculation

Similar Questions

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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

Select Answer:

Visualized Solution

Visualizing the Circle and Lines

The Normal Property

Substituting the Center

Tangency Condition

Distance to Tangent 1

Simplifying Distance 1

Distance to Tangent 2

Equating the Distances

Solving the Absolute Value (Case 1)

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

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

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

Checking Constraints

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Anchor of the Normal

The Tangency Constraint

The Algebraic Dance

Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The common tangent to the circles x2+y2=4 and x2+y2+6x+8y−24=0 also passes through the point :-

Finding $β$

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Calculating Radius $r$

The number of common tangents to the circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} - 6 x - 8 y = 24$ is

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