Extremities of a diagonal of a rectangle are (0, 0) and (4, 3). The equations of the tangents to the circumcircle of the rectangle which are parallel to the diagonal, are

Options

A. `16x+8ypm25=0`
B. `6x-8ypm25=0`
C. `8+6ypm25=0`
D. none of these

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Identify the coordinates of the rectangle's diagonal
The extremities of the diagonal of the rectangle are given as points \( A(0, 0) \) and \( B(4, 3) \).

### Step 2: Find the center and radius of the circumcircle
The center of the circumcircle (which is also the midpoint of the diagonal) can be calculated using the midpoint formula:
\[ C\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = C\left( \frac{0 + 4}{2}, \frac{0 + 3}{2} \right) = C(2, \frac{3}{2}) \]

Next, we calculate the diameter of the circle using the distance formula:
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(4 - 0)^2 + (3 - 0)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]
The radius \( r \) is half of the diameter:
\[ r = \frac{5}{2} \]

### Step 3: Write the equation of the circumcircle
The equation of a circle with center \( (h, k) \) and radius \( r \) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting the values:
\[ (x - 2)^2 + \left(y - \frac{3}{2}\right)^2 = \left(\frac{5}{2}\right)^2 \]
This simplifies to:
\[ (x - 2)^2 + \left(y - \frac{3}{2}\right)^2 = \frac{25}{4} \]

### Step 4: Find the slope of the diagonal
The slope \( m \) of the diagonal \( AB \) is calculated as:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 0}{4 - 0} = \frac{3}{4} \]

### Step 5: Write the equations of the tangents parallel to the diagonal
The equation of the tangent line to the circle at point \( (x_1, y_1) \) with slope \( m \) is given by:
\[ y - k = m(x - h) \pm r \sqrt{1 + m^2} \]
Substituting \( h = 2 \), \( k = \frac{3}{2} \), \( m = \frac{3}{4} \), and \( r = \frac{5}{2} \):
\[ y - \frac{3}{2} = \frac{3}{4}(x - 2) \pm \frac{5}{2} \sqrt{1 + \left(\frac{3}{4}\right)^2} \]
Calculating \( \sqrt{1 + \left(\frac{3}{4}\right)^2} \):
\[ \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4} \]
Thus, the equation becomes:
\[ y - \frac{3}{2} = \frac{3}{4}(x - 2) \pm \frac{5}{2} \cdot \frac{5}{4} \]
This simplifies to:
\[ y - \frac{3}{2} = \frac{3}{4}(x - 2) \pm \frac{25}{8} \]

### Step 6: Simplify the tangent equations
1. For the positive case:
\[ y - \frac{3}{2} = \frac{3}{4}(x - 2) + \frac{25}{8} \]
Rearranging gives:
\[ 8y - 12 = 6x - 12 + 25 \implies 6x - 8y + 25 = 0 \]

2. For the negative case:
\[ y - \frac{3}{2} = \frac{3}{4}(x - 2) - \frac{25}{8} \]
Rearranging gives:
\[ 8y - 12 = 6x - 12 - 25 \implies 6x - 8y - 25 = 0 \]

### Final Result
The equations of the tangents to the circumcircle of the rectangle which are parallel to the diagonal are:
\[ 6x - 8y + 25 = 0 \quad \text{and} \quad 6x - 8y - 25 = 0 \]