The lines 3x - 4y + 4 = 0 and 6x - 8y - 7 = 0 are tangents to the same circle. The radius of this ci

Q: Wait, why do we reject the negative root here? Or why can't we just use energy conservation?

This is a pure geometry problem, so energy conservation doesn't apply! As for the negative root, the distance between two lines is a physical length, which must always be positive. That is why we use the absolute value $|c_1 - c_2|$ in our formula.

Q: Why did we have to divide the second equation by 2? Can't we just use the coefficients as they are?

Great question! The standard distance formula $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$ only works when the $x$ and $y$ coefficients ($a$ and $b$) are identical in both equations. By dividing the second equation by $2$, we normalize it so that the $a$ and $b$ values match the first equation, allowing the formula to work correctly.

Q: Is the distance between the lines the radius or the diameter?

Think about the circle 'sandwiched' between the two parallel lines. The line segment connecting the two points of tangency passes through the center and is perpendicular to both lines. This segment is the diameter of the circle. So, the distance between the lines is the diameter, and you must divide by $2$ to get the radius.

Question

The lines $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$ are tangents to the same circle. The radius of this circle is .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, aspiring mathematician! Today, we are going to unravel a beautiful problem in coordinate geometry. It is not just about plugging numbers into a formula; it is about visualizing the elegant dance between lines and circles.

Imagine you are standing on a vast coordinate plane. You see two lines, $L_1: 3x - 4y + 4 = 0$ and $L_2: 6x - 8y - 7 = 0$. You are told that a single circle is tangent to both.

### Phase 1: The Hidden Parallelism

First, let us look at the slopes. For any line $ax + by + c = 0$, the slope is given by $-\frac{a}{b}$. For our first line, $L_1$, the slope is $-\frac{3}{-4} = \frac{3}{4}$.

Now, look at $L_2$. The slope is $-\frac{6}{-8}$, which simplifies perfectly to $\frac{3}{4}$.

Do you see it? The slopes are identical! This means our two lines are perfectly parallel.

If a circle is trapped between two parallel lines, it must be touching both of them simultaneously. This creates a 'sandwich' effect. The distance between these two lines is not just some random gap; it is the diameter of the circle!

### Phase 2: The Normalization Trap

Now, here is where many students stumble. We want to use the distance formula for parallel lines:

$$d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$$

But wait! This formula only works if the coefficients of $x$ and $y$ are identical in both equations. Currently, $L_1$ has coefficients $3$ and $-4$, while $L_2$ has $6$ and $-8$.

We must normalize them. Let us divide the entire equation of $L_2$ by $2$. This gives us $3x - 4y - \frac{7}{2} = 0$. Now, the coefficients match perfectly.

### Phase 3: The Elegant Calculation

With our equations $3x - 4y + 4 = 0$ and $3x - 4y - \frac{7}{2} = 0$, we can identify our constants: $c_1 = 4$ and $c_2 = -\frac{7}{2}$. The coefficients are $a = 3$ and $b = -4$.

Plugging these into our distance formula:

$$d = \frac{|4 - (-\frac{7}{2})|}{\sqrt{3^2 + (-4)^2}}$$

Look at the numerator: $4 + \frac{7}{2} = \frac{8}{2} + \frac{7}{2} = \frac{15}{2}$.

Look at the denominator: $\sqrt{9 + 16} = \sqrt{25} = 5$.

So, the diameter $d$ is:

$$d = \frac{15/2}{5} = \frac{15}{10} = \frac{3}{2}$$

### The Final Step

We have found the diameter $d = \frac{3}{2}$. But do not stop yet! The question asks for the radius. The radius $r$ is simply half of the diameter:

$$r = \frac{d}{2} = \frac{3/2}{2} = \frac{3}{4} = 0.75$$

And there it is! By visualizing the geometry and carefully normalizing our equations, we have arrived at the answer. The final radius is **$0.75$**.

Animated Solution for Mathematics - Circles: The lines $3 x - 4 y + 4 = 0$ and $6 x - 8 y - 7 = 0$ are tangents to the same circle. The radius of this circle is .........

Enter Numerical Value:

Visualized Solution

Visualizing the Tangents

Checking for Parallelism

Geometric Relationship

Normalizing the Equations

Distance Formula

Substituting Values

Simplifying the Expression

Calculating the Diameter

Finding the Radius

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Phase 1

Phase 2

Phase 3

The Final Step

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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; } The lines 3x−4y+4=0 and 6x−8y−7=0 are tangents to the same circle. The radius of this circle is .........