Equation to the circles which pass through the point (2,3) and cut off equal chords of length 6 units along the lines y-x-1=0 and y+x-5=0 is

Options

A. `(x-3)^2+(y-2sqrt3)^2=(3sqrt2)^2`
B. `(x-2)^2+(y-3-3sqrt2)^2=(3sqrt2)^2`
C. `(x-2sqrt3)^2 + (y-3)^2 =(3sqrt2)^2`
D. `none of these

Answer

To find the equation of the circles that pass through the point (2, 3) and cut off equal chords of length 6 units along the lines \( y - x - 1 = 0 \) and \( y + x - 5 = 0 \), we can follow these steps:

### Step 1: Find the intersection point of the two lines
The equations of the lines are:
1. \( y - x - 1 = 0 \) (or \( y = x + 1 \))
2. \( y + x - 5 = 0 \) (or \( y = 5 - x \))

To find the intersection, we can set the two equations equal to each other:
\[ x + 1 = 5 - x \]
Adding \( x \) to both sides gives:
\[ 2x + 1 = 5 \]
Subtracting 1 from both sides:
\[ 2x = 4 \]
Dividing by 2:
\[ x = 2 \]
Substituting \( x = 2 \) back into one of the equations to find \( y \):
\[ y = 2 + 1 = 3 \]
Thus, the intersection point is \( (2, 3) \).

### Step 2: Determine the slopes of the lines
The slope of the first line \( y - x - 1 = 0 \) is 1, and the slope of the second line \( y + x - 5 = 0 \) is -1. Since the product of their slopes is -1, the lines are perpendicular.

### Step 3: Use the property of the circle
Since the chords cut off by the lines are equal and the lines are perpendicular, the center of the circle lies on the line that bisects the angle between the two lines. The length of each chord is given as 6 units.

### Step 4: Find the radius of the circle
Using the property of circles, if two equal chords of length \( L \) are cut off by two perpendicular lines, we can find the radius \( r \) of the circle using the formula:
\[ r = \sqrt{\left(\frac{L}{2}\right)^2 + d^2} \]
where \( d \) is the distance from the center of the circle to the chord.

Since the length of the chord \( L = 6 \):
\[ \frac{L}{2} = 3 \]
The distance \( d \) from the center to the chord can be found using the fact that the center lies on the angle bisector. For equal chords, we can assume \( d \) to be \( r \) (as the center is equidistant from both chords).

### Step 5: Calculate the radius
Using the Pythagorean theorem, we have:
\[ r^2 = 3^2 + 3^2 = 9 + 9 = 18 \]
Thus,
\[ r = 3\sqrt{2} \]

### Step 6: Write the equation of the circle
The general equation of a circle is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \( (h, k) \) is the center of the circle. Since the circle passes through the point \( (2, 3) \) and has radius \( 3\sqrt{2} \), we can substitute \( h = 2 \), \( k = 3 \), and \( r = 3\sqrt{2} \):
\[ (x - 2)^2 + (y - 3)^2 = (3\sqrt{2})^2 \]
\[ (x - 2)^2 + (y - 3)^2 = 18 \]

### Final Answer
The equation of the circles is:
\[ (x - 2)^2 + (y - 3)^2 = 18 \]