On the line joining the points A (0,4) and B (3,0), a square ABCD is constructed on the side of the line away from the origin. Equation of the circle having centre at C and touching the axis of x is

Options

A. `x ^(2) + y ^(2) - 14 x - 6y + 49 =0`
B. `x ^(2) + y ^(2) - 14 x- 6y + 9=0`
C. `x ^(2) + y ^(2) - 6x - 14 y + 49 =0`
D. `x ^(2) + y ^(2) - 6x - 14 y + 9=0`

Answer

To solve the problem, we need to follow these steps:

### Step 1: Find the slope of line AB
The points A and B are given as A(0, 4) and B(3, 0). The slope \( m \) of line AB can be calculated using the formula:

\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 4}{3 - 0} = \frac{-4}{3} \]

### Step 2: Find the equation of line AB
Using point-slope form of the line equation, \( y - y_1 = m(x - x_1) \), we can write the equation of line AB:

\[ y - 4 = -\frac{4}{3}(x - 0) \]

This simplifies to:

\[ y = -\frac{4}{3}x + 4 \]

### Step 3: Find the coordinates of point C
Since we are constructing a square ABCD on the side of the line away from the origin, we need to find the coordinates of point C. The line perpendicular to AB will have a slope that is the negative reciprocal of \( m \):

\[ m_{perpendicular} = \frac{3}{4} \]

Using point A(0, 4) to find the equation of the line through A that is perpendicular to AB:

\[ y - 4 = \frac{3}{4}(x - 0) \]

This simplifies to:

\[ y = \frac{3}{4}x + 4 \]

### Step 4: Find the intersection point D
The coordinates of point D can be found by solving the equations of the two lines (line AB and the perpendicular line through A). We set the equations equal to each other:

\[ -\frac{4}{3}x + 4 = \frac{3}{4}x + 4 \]

Solving for \( x \):

\[ -\frac{4}{3}x = \frac{3}{4}x \]

Multiplying through by 12 to eliminate fractions:

\[ -16x = 9x \]

\[ -25x = 0 \implies x = 0 \]

Substituting \( x = 0 \) back into either line equation gives \( y = 4 \). Thus, point D coincides with point A, which is not possible. We need to find the coordinates of point C directly.

### Step 5: Calculate the coordinates of C
The distance between points A and B can be calculated using the distance formula:

\[ AB = \sqrt{(3 - 0)^2 + (0 - 4)^2} = \sqrt{9 + 16} = 5 \]

The side of the square will be equal to the distance AB, which is 5. The coordinates of point C can be found by moving along the perpendicular direction from point A:

Let the coordinates of C be \( (h, k) \). The distance from A to C will also be 5, and since we are moving perpendicular, we can use the direction vector of the perpendicular line.

### Step 6: Find the center of the circle
The center of the circle is at point C, which is at a distance of 5 units from A. The coordinates of C can be derived from the direction of the perpendicular slope.

### Step 7: Circle equation
The equation of a circle with center at C(h, k) and radius r is given by:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

Since the circle touches the x-axis, the radius \( r \) will be equal to the y-coordinate of point C (k).

### Final Equation
Thus, the equation of the circle can be expressed as:

\[ (x - h)^2 + (y - k)^2 = k^2 \]