The coordinates of the points on the circles `x^(2) +y^(2) =9` which are at a distance of 5 units from the line `3x + 4y = 25` are `:`

Options

A. (a) `(-2,3)` and `( 3,-2)`
B. (b) `(-3,1)` and `( 3,-1)`
C. (c) `((12)/( 5) ,( -9)/( 5))`,`((-12)/( 5) ,( 9)/( 5))`
D. (d) None of these

Answer

To solve the problem, we need to find the coordinates of the points on the circle defined by the equation \(x^2 + y^2 = 9\) that are at a distance of 5 units from the line given by the equation \(3x + 4y = 25\).

### Step 1: Understand the Circle and Line
The equation of the circle \(x^2 + y^2 = 9\) represents a circle centered at the origin (0, 0) with a radius of 3 units. The line \(3x + 4y = 25\) can be rewritten in slope-intercept form or used directly to find the distance from points to the line.

**Hint:** The distance from a point \((a, b)\) to a line \(Ax + By + C = 0\) is given by the formula:
\[ D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

### Step 2: Find the Distance from the Line
The line can be rewritten as:
\[ 3x + 4y - 25 = 0 \]
Here, \(A = 3\), \(B = 4\), and \(C = -25\).

We want the distance \(D\) from a point \((a, b)\) on the circle to this line to be equal to 5 units. Thus, we have:
\[ D = \frac{|3a + 4b - 25|}{\sqrt{3^2 + 4^2}} = 5 \]
Calculating the denominator:
\[ \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]
So the equation simplifies to:
\[ |3a + 4b - 25| = 25 \]

### Step 3: Set Up the Equations
This absolute value equation gives us two cases:
1. \(3a + 4b - 25 = 25\)
2. \(3a + 4b - 25 = -25\)

**Case 1:**
\[ 3a + 4b = 50 \quad \text{(Equation 1)} \]

**Case 2:**
\[ 3a + 4b = 0 \quad \text{(Equation 2)} \]

### Step 4: Solve Each Case with the Circle Equation
Now we will solve each case along with the circle equation \(a^2 + b^2 = 9\).

**Case 1:**
From \(3a + 4b = 50\):
- Express \(b\) in terms of \(a\):
\[ 4b = 50 - 3a \implies b = \frac{50 - 3a}{4} \]
Substituting into the circle equation:
\[ a^2 + \left(\frac{50 - 3a}{4}\right)^2 = 9 \]
Expanding and simplifying:
\[ a^2 + \frac{(50 - 3a)^2}{16} = 9 \]
Multiply through by 16 to eliminate the fraction:
\[ 16a^2 + (50 - 3a)^2 = 144 \]
Expanding \((50 - 3a)^2\):
\[ 16a^2 + 2500 - 300a + 9a^2 = 144 \]
Combine like terms:
\[ 25a^2 - 300a + 2356 = 0 \]
This quadratic can be solved using the quadratic formula.

**Case 2:**
From \(3a + 4b = 0\):
- Express \(b\) in terms of \(a\):
\[ 4b = -3a \implies b = -\frac{3a}{4} \]
Substituting into the circle equation:
\[ a^2 + \left(-\frac{3a}{4}\right)^2 = 9 \]
Expanding:
\[ a^2 + \frac{9a^2}{16} = 9 \]
Multiply through by 16:
\[ 16a^2 + 9a^2 = 144 \]
Combine like terms:
\[ 25a^2 = 144 \implies a^2 = \frac{144}{25} \implies a = \pm \frac{12}{5} \]
Substituting back to find \(b\):
\[ b = -\frac{3(\pm \frac{12}{5})}{4} = \mp \frac{9}{5} \]

### Final Coordinates
Thus, the coordinates from Case 2 are:
1. \(\left(\frac{12}{5}, -\frac{9}{5}\right)\)
2. \(\left(-\frac{12}{5}, \frac{9}{5}\right)\)

### Conclusion
The coordinates of the points on the circle that are at a distance of 5 units from the line \(3x + 4y = 25\) are:
\[ \left(\frac{12}{5}, -\frac{9}{5}\right) \quad \text{and} \quad \left(-\frac{12}{5}, \frac{9}{5}\right) \]