The equation of the circle which cuts the circel `x ^(2) +y ^(2) = a ^(2)` ortiogonally at the point `((a)/(sqrt2), (a)/(sqrt2))` such that slope of their common chord is -1 is

Options

A. `x ^(2) +y ^(2) + 2sqrt2 x _ 2 sqrt2 y +a ^(2) =0`
B. `x ^(2) + y ^(2) -a sqrt2 x -a sqrt2 y -a ^(2) =0`
C. `x ^(2) + y ^(2) + sqrt2x + sqrt2y + a^(2) =0`
D. `x ^(2) +y ^(2) -a sqrt2 x -a sqrt2y +a ^(2) =0`

Answer

To find the equation of the circle that intersects the circle \( x^2 + y^2 = a^2 \) orthogonally at the point \(\left(\frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}}\right)\) and has a common chord with a slope of -1, we can follow these steps:

### Step 1: General Equation of the Circle
We start with the general equation of a circle:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
where \(g\), \(f\), and \(c\) are constants to be determined.

### Step 2: Condition for Orthogonality
For two circles to intersect orthogonally, the following condition must hold:
\[ 2g_1g_2 + 2f_1f_2 = c_1 + c_2 \]
Here, we have the first circle \(x^2 + y^2 = a^2\) which can be rewritten as:
\[ x^2 + y^2 - a^2 = 0 \]
Thus, \(g_1 = 0\), \(f_1 = 0\), and \(c_1 = -a^2\). For our circle, we denote \(g_2 = g\), \(f_2 = f\), and \(c_2 = c\). The condition becomes:
\[ 0 + 0 = -a^2 + c \implies c = a^2 \]

### Step 3: Substitute \(c\) into the Circle Equation
Now, substituting \(c = a^2\) into the general circle equation gives:
\[ x^2 + y^2 + 2gx + 2fy + a^2 = 0 \]

### Step 4: Find the Equation of the Common Chord
The equation of the common chord can be derived from the two circles. The common chord is given by:
\[ S_1 - S_2 = 0 \]
where \(S_1\) is the equation of the first circle and \(S_2\) is the equation of our circle. Thus, we have:
\[ (x^2 + y^2 - a^2) - (x^2 + y^2 + 2gx + 2fy + a^2) = 0 \]
This simplifies to:
\[ -2gx - 2fy - 2a^2 = 0 \implies 2gx + 2fy + 2a^2 = 0 \]
Dividing through by 2 gives:
\[ gx + fy + a^2 = 0 \]

### Step 5: Slope of the Common Chord
The slope of the common chord is given by:
\[ \text{slope} = -\frac{g}{f} \]
We are given that the slope is -1, so:
\[ -\frac{g}{f} = -1 \implies g = f \]

### Step 6: Substitute \(g = f\) into the Circle Equation
Now substituting \(g = f\) into the circle equation:
\[ x^2 + y^2 + 2gx + 2gy + a^2 = 0 \]
This can be rewritten as:
\[ x^2 + y^2 + 2g(x + y) + a^2 = 0 \]

### Step 7: Substitute the Point of Intersection
The circle intersects at the point \(\left(\frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}}\right)\). Substituting this point into the equation:
\[ \left(\frac{a}{\sqrt{2}}\right)^2 + \left(\frac{a}{\sqrt{2}}\right)^2 + 2g\left(\frac{a}{\sqrt{2}} + \frac{a}{\sqrt{2}}\right) + a^2 = 0 \]
This simplifies to:
\[ \frac{a^2}{2} + \frac{a^2}{2} + 2g\left(\frac{2a}{\sqrt{2}}\right) + a^2 = 0 \]
\[ a^2 + 2g\left(\frac{2a}{\sqrt{2}}\right) + a^2 = 0 \]
\[ 2a^2 + \frac{4ag}{\sqrt{2}} = 0 \]
\[ \frac{4ag}{\sqrt{2}} = -2a^2 \implies g = -\frac{a\sqrt{2}}{4} \]

### Step 8: Final Equation of the Circle
Substituting \(g\) back into the circle equation:
\[ x^2 + y^2 - \frac{a\sqrt{2}}{2}(x + y) + a^2 = 0 \]
This can be rewritten as:
\[ x^2 + y^2 - \sqrt{2}a x - \sqrt{2}a y + a^2 = 0 \]

### Final Answer
The final equation of the circle is:
\[ x^2 + y^2 - \sqrt{2}a x - \sqrt{2}a y + a^2 = 0 \]