The area (in sq. units) of the circle touching the line `x+y=4` at (1, 3) and intersecting `x^(2)+y^(2)=4` orthogonally is equal to

Options

A. `(9pi)/(8)`
B. `(7pi)/(8)`
C. `(5pi)/(4)`
D. `(4pi)/(3)`

Answer

To solve the problem, we need to find the area of a circle that touches the line \(x + y = 4\) at the point \((1, 3)\) and intersects the circle defined by \(x^2 + y^2 = 4\) orthogonally.

### Step-by-Step Solution:

1. **Identify the Circle's General Equation**:
The circle touches the line \(x + y = 4\) at the point \((1, 3)\). The general equation of a circle can be written as:
\[ (x - 1)^2 + (y - 3)^2 + \lambda (x + y - 4) = 0 \]
Here, \(\lambda\) is a parameter that we will determine later.

2. **Expand the Circle's Equation**:
Expanding the equation gives:
\[ (x^2 - 2x + 1) + (y^2 - 6y + 9) + \lambda (x + y - 4) = 0 \]
This simplifies to:
\[ x^2 + y^2 + (\lambda - 2)x + (\lambda - 6)y + (10 - 4\lambda) = 0 \]

3. **Identify the Second Circle**:
The second circle is given by:
\[ x^2 + y^2 = 4 \]
This can be rewritten in the standard form:
\[ x^2 + y^2 + 0x + 0y - 4 = 0 \]
Here, we have \(G_2 = 0\), \(F_2 = 0\), and \(C_2 = -4\).

4. **Use the Orthogonality Condition**:
For two circles to intersect orthogonally, the condition is:
\[ 2G_1G_2 + 2F_1F_2 = C_1 + C_2 \]
From our first circle, we have:
- \(G_1 = \frac{\lambda - 2}{2}\)
- \(F_1 = \frac{\lambda - 6}{2}\)
- \(C_1 = 10 - 4\lambda\)

Plugging in the values:
\[ 2 \left(\frac{\lambda - 2}{2}\right)(0) + 2 \left(\frac{\lambda - 6}{2}\right)(0) = (10 - 4\lambda) - 4 \]
This simplifies to:
\[ 0 = 6 - 4\lambda \]
Solving for \(\lambda\):
\[ 4\lambda = 6 \implies \lambda = \frac{3}{2} \]

5. **Substitute \(\lambda\) Back into the Circle's Equation**:
Substitute \(\lambda = \frac{3}{2}\) into the circle's equation:
\[ x^2 + y^2 + \left(\frac{3}{2} - 2\right)x + \left(\frac{3}{2} - 6\right)y + (10 - 4 \cdot \frac{3}{2}) = 0 \]
This simplifies to:
\[ x^2 + y^2 - \frac{1}{2}x - \frac{9}{2}y + 4 = 0 \]

6. **Complete the Square**:
Rearranging gives:
\[ x^2 - \frac{1}{2}x + y^2 - \frac{9}{2}y + 4 = 0 \]
Completing the square for \(x\) and \(y\):
\[ \left(x - \frac{1}{4}\right)^2 - \frac{1}{16} + \left(y - \frac{9}{4}\right)^2 - \frac{81}{16} + 4 = 0 \]
This leads to:
\[ \left(x - \frac{1}{4}\right)^2 + \left(y - \frac{9}{4}\right)^2 = \frac{82}{16} - 4 \]
Simplifying gives:
\[ \left(x - \frac{1}{4}\right)^2 + \left(y - \frac{9}{4}\right)^2 = \frac{18}{16} = \frac{9}{8} \]

7. **Find the Radius**:
The radius squared \(r^2 = \frac{9}{8}\), thus:
\[ r = \sqrt{\frac{9}{8}} = \frac{3}{2\sqrt{2}} = \frac{3\sqrt{2}}{4} \]

8. **Calculate the Area**:
The area \(A\) of the circle is given by:
\[ A = \pi r^2 = \pi \cdot \frac{9}{8} = \frac{9\pi}{8} \]

### Final Answer:
The area of the circle is \(\frac{9\pi}{8}\) square units.