The lines joining the origin to the points of intersection of the line `4x+3y=24` with the circle `(x-3)^(2)+(y-4)^(2)=25` are

Options

A. coincident
B. perpendicular
C. equally inclined to x - axis
D. None of these

Answer

To solve the problem of finding the lines joining the origin to the points of intersection of the line \(4x + 3y = 24\) with the circle \((x - 3)^2 + (y - 4)^2 = 25\), we will follow these steps:

### Step 1: Identify the equations
We have the line:
\[ 4x + 3y = 24 \]
And the circle:
\[ (x - 3)^2 + (y - 4)^2 = 25 \]

### Step 2: Express \(y\) in terms of \(x\) from the line equation
From the line equation, we can express \(y\) in terms of \(x\):
\[ 3y = 24 - 4x \implies y = \frac{24 - 4x}{3} \]

### Step 3: Substitute \(y\) into the circle equation
Now, substitute \(y\) into the circle equation:
\[ (x - 3)^2 + \left(\frac{24 - 4x}{3} - 4\right)^2 = 25 \]
First, simplify \(\frac{24 - 4x}{3} - 4\):
\[ \frac{24 - 4x}{3} - 4 = \frac{24 - 4x - 12}{3} = \frac{12 - 4x}{3} \]
Now substitute this back into the circle equation:
\[ (x - 3)^2 + \left(\frac{12 - 4x}{3}\right)^2 = 25 \]

### Step 4: Expand and simplify the equation
Expanding \((x - 3)^2\):
\[ (x - 3)^2 = x^2 - 6x + 9 \]
Expanding \(\left(\frac{12 - 4x}{3}\right)^2\):
\[ \left(\frac{12 - 4x}{3}\right)^2 = \frac{(12 - 4x)^2}{9} = \frac{144 - 96x + 16x^2}{9} \]
Now, combine these two expressions:
\[ x^2 - 6x + 9 + \frac{144 - 96x + 16x^2}{9} = 25 \]
Multiply through by 9 to eliminate the fraction:
\[ 9(x^2 - 6x + 9) + 144 - 96x + 16x^2 = 225 \]
This simplifies to:
\[ 9x^2 - 54x + 81 + 144 - 96x + 16x^2 = 225 \]
Combine like terms:
\[ (9x^2 + 16x^2) + (-54x - 96x) + (81 + 144 - 225) = 0 \]
\[ 25x^2 - 150x = 0 \]

### Step 5: Factor the equation
Factoring out \(25x\):
\[ 25x(x - 6) = 0 \]
This gives us the solutions:
\[ x = 0 \quad \text{or} \quad x = 6 \]

### Step 6: Find corresponding \(y\) values
Substituting \(x = 0\) into the line equation:
\[ 4(0) + 3y = 24 \implies 3y = 24 \implies y = 8 \quad \Rightarrow (0, 8) \]
Substituting \(x = 6\) into the line equation:
\[ 4(6) + 3y = 24 \implies 24 + 3y = 24 \implies 3y = 0 \implies y = 0 \quad \Rightarrow (6, 0) \]

### Step 7: Determine the lines from the origin
The points of intersection are \((0, 8)\) and \((6, 0)\). The lines from the origin to these points are:
1. From \((0, 0)\) to \((0, 8)\) which is vertical.
2. From \((0, 0)\) to \((6, 0)\) which is horizontal.

### Conclusion
The lines joining the origin to the points of intersection of the line and the circle are perpendicular to each other.