If the area of a quadrilateral formed by the tangents from the origin to `x^(2) + y^(2) + 6x - 10y + lambda =0` and the pair of radii at the points of contact of these tangents to the circle, is 8 `cm^(2)`, then the value of `lambda` must be

Options

A. 2
B. 4
C. 16
D. 32

Answer

To solve the problem, we need to find the value of \(\lambda\) for which the area of the quadrilateral formed by the tangents from the origin to the circle defined by the equation \(x^2 + y^2 + 6x - 10y + \lambda = 0\) and the pair of radii at the points of contact of these tangents is \(8 \, \text{cm}^2\).

### Step 1: Identify the center and radius of the circle

The standard form of a circle is given by:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From the given equation, we can identify:
- \(g = 3\) (since \(6x = 2gx\))
- \(f = -5\) (since \(-10y = 2fy\))
- \(c = \lambda\)

The center of the circle \((h, k)\) can be found using:
\[ (h, k) = (-g, -f) = (-3, 5) \]

### Step 2: Calculate the radius of the circle

The radius \(r\) of the circle can be calculated using the formula:
\[ r = \sqrt{g^2 + f^2 - c} \]
Substituting the values we have:
\[ r = \sqrt{3^2 + (-5)^2 - \lambda} = \sqrt{9 + 25 - \lambda} = \sqrt{34 - \lambda} \]

### Step 3: Find the length of the tangents from the origin

The length of the tangents from the origin to the circle is given by:
\[ L = \sqrt{S_1} \]
where \(S_1\) is obtained by substituting the coordinates of the point (origin) into the circle's equation:
\[ S_1 = 0^2 + 0^2 + 6(0) - 10(0) + \lambda = \lambda \]
Thus, the length of the tangents is:
\[ L = \sqrt{\lambda} \]

### Step 4: Calculate the area of triangle OAB

The area \(A\) of triangle OAB formed by the origin and the points of tangency can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is the length of the tangent \(L = \sqrt{\lambda}\) and the height is the radius \(r = \sqrt{34 - \lambda}\):
\[ A = \frac{1}{2} \times \sqrt{\lambda} \times \sqrt{34 - \lambda} \]

### Step 5: Area of quadrilateral OABC

The area of quadrilateral OABC is twice the area of triangle OAB:
\[ \text{Area}_{OABC} = 2A = \sqrt{\lambda} \times \sqrt{34 - \lambda} \]
We are given that this area equals \(8 \, \text{cm}^2\):
\[ \sqrt{\lambda} \times \sqrt{34 - \lambda} = 8 \]

### Step 6: Square both sides

Squaring both sides gives:
\[ \lambda(34 - \lambda) = 64 \]
Expanding this, we get:
\[ 34\lambda - \lambda^2 = 64 \]
Rearranging gives:
\[ \lambda^2 - 34\lambda + 64 = 0 \]

### Step 7: Solve the quadratic equation

Using the quadratic formula:
\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 1\), \(b = -34\), and \(c = 64\):
\[ \lambda = \frac{34 \pm \sqrt{(-34)^2 - 4 \cdot 1 \cdot 64}}{2 \cdot 1} \]
Calculating the discriminant:
\[ \lambda = \frac{34 \pm \sqrt{1156 - 256}}{2} = \frac{34 \pm \sqrt{900}}{2} = \frac{34 \pm 30}{2} \]
This gives us two solutions:
\[ \lambda = \frac{64}{2} = 32 \quad \text{and} \quad \lambda = \frac{4}{2} = 2 \]

### Conclusion

The possible values of \(\lambda\) are \(2\) and \(32\).