Two lines through `(2, 3)` from which the circle `x^2+y^2 =25` intercepts chords of length `8` units have equations
(A) `2x+3y=13`, `x+5y=17`
(B) `y= 3`, `12x+5y=39`
(C) `x=2`, `9x -11y=51`
(D) `y=0`, `12x+5y=39`

Options

A. `y=3`
B. `12x+5y=39`
C. `x=2`
D. `9x-11y=51`

Explanation

To solve the problem of finding the equations of two lines through the point (2, 3) from which the circle \(x^2 + y^2 = 25\) intercepts chords of length 8 units, we can follow these steps:

### Step 1: Understand the Circle and Chord Length
The given circle is \(x^2 + y^2 = 25\), which has a center at (0, 0) and a radius of 5 (since \(\sqrt{25} = 5\)). The length of the chord intercepted by the lines is given as 8 units.

**Hint:** Remember that the length of the chord can be related to the perpendicular distance from the center of the circle to the chord.

### Step 2: Calculate the Half Chord Length
Since the full length of the chord is 8 units, the half-length of the chord is:
\[ \text{Half Chord Length} = \frac{8}{2} = 4 \text{ units} \]

**Hint:** The half-length of the chord is used to find the distance from the center of the circle to the chord.

### Step 3: Use the Perpendicular Distance Formula
The distance \(d\) from the center of the circle (0, 0) to the chord can be calculated using the Pythagorean theorem:
\[ d = \sqrt{r^2 - (\text{Half Chord Length})^2} \]
Substituting the values:
\[ d = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3 \]

**Hint:** This distance \(d\) is crucial for finding the slope of the lines.

### Step 4: Set Up the Equation of the Line
Let the slope of the line be \(m\). The equation of the line passing through the point (2, 3) can be written as:
\[ y - 3 = m(x - 2) \]
Rearranging gives:
\[ mx - y - 2m + 3 = 0 \]

**Hint:** This form will help us find the distance from the center of the circle to the line.

### Step 5: Calculate the Distance from the Center to the Line
The distance \(d\) from the center of the circle (0, 0) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|A(0) + B(0) + C|}{\sqrt{A^2 + B^2}} = \frac{|C|}{\sqrt{A^2 + B^2}} \]
For our line, \(A = m\), \(B = -1\), and \(C = -2m + 3\):
\[ d = \frac{|-2m + 3|}{\sqrt{m^2 + 1}} = 3 \]

**Hint:** Set this distance equal to the value calculated earlier (3) to find the slope.

### Step 6: Solve the Equation
Setting the equation:
\[ \frac{| -2m + 3 |}{\sqrt{m^2 + 1}} = 3 \]
Squaring both sides gives:
\[ (-2m + 3)^2 = 9(m^2 + 1) \]
Expanding both sides:
\[ 4m^2 - 12m + 9 = 9m^2 + 9 \]
Rearranging gives:
\[ 5m^2 + 12m = 0 \]
Factoring out \(m\):
\[ m(5m + 12) = 0 \]
Thus, \(m = 0\) or \(m = -\frac{12}{5}\).

**Hint:** Each slope corresponds to a different line equation.

### Step 7: Find the Equations of the Lines
1. For \(m = 0\):
\[ y - 3 = 0 \implies y = 3 \]

2. For \(m = -\frac{12}{5}\):
\[ y - 3 = -\frac{12}{5}(x - 2) \]
Rearranging gives:
\[ 12x + 5y = 39 \]

**Hint:** These two lines are the final answers.

### Final Answer
The two lines are:
- \(y = 3\)
- \(12x + 5y = 39\)