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, we need to find the equations of the tangents to the circumcircle of a rectangle, which are parallel to the diagonal defined by the points (0, 0) and (4, 3).

### Step-by-Step Solution:

1. **Identify the Coordinates of the Rectangle's Diagonal:**
The extremities of the diagonal are given as \( A(0, 0) \) and \( B(4, 3) \).

2. **Find the Midpoint of the Diagonal:**
The midpoint \( M \) of the diagonal can be calculated using the midpoint formula:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{0 + 4}{2}, \frac{0 + 3}{2} \right) = (2, 1.5) \]

3. **Determine the Length of the Diagonal:**
The length of the diagonal \( AB \) can be calculated using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(4 - 0)^2 + (3 - 0)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

4. **Find the Radius of the Circumcircle:**
The radius \( R \) of the circumcircle is half the length of the diagonal:
\[ R = \frac{d}{2} = \frac{5}{2} = 2.5 \]

5. **Write the Equation of the Circumcircle:**
The equation of the circumcircle with center \( M(2, 1.5) \) and radius \( R = 2.5 \) is given by:
\[ (x - 2)^2 + (y - 1.5)^2 = (2.5)^2 \]
Simplifying this, we have:
\[ (x - 2)^2 + (y - 1.5)^2 = 6.25 \]

6. **Determine the Slope of the Diagonal:**
The slope \( m \) of the diagonal \( AB \) is calculated as:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 0}{4 - 0} = \frac{3}{4} \]

7. **Find the Equation of the Tangents Parallel to the Diagonal:**
The slope of the tangents will also be \( \frac{3}{4} \). The general equation of a line with slope \( m \) passing through a point \( (x_0, y_0) \) is:
\[ y - y_0 = m(x - x_0) \]
Here, \( (x_0, y_0) \) will be the center of the circle \( (2, 1.5) \).

Thus, the equation of the tangent lines will be:
\[ y - 1.5 = \frac{3}{4}(x - 2) + c \]
where \( c \) can be \( \pm \sqrt{R^2 - m^2} \).

8. **Calculate the Value of \( c \):**
First, calculate \( R^2 - m^2 \):
\[ R^2 = \left( \frac{5}{2} \right)^2 = 6.25 \]
\[ m^2 = \left( \frac{3}{4} \right)^2 = \frac{9}{16} \]
\[ R^2 - m^2 = 6.25 - \frac{9}{16} = \frac{100}{16} - \frac{9}{16} = \frac{91}{16} \]
Thus, \( c = \pm \sqrt{\frac{91}{16}} = \pm \frac{\sqrt{91}}{4} \).

9. **Final Tangent Equations:**
The equations of the tangents are:
\[ y - 1.5 = \frac{3}{4}(x - 2) + \frac{\sqrt{91}}{4} \]
\[ y - 1.5 = \frac{3}{4}(x - 2) - \frac{\sqrt{91}}{4} \]