Tangent drawn from the point `(a ,3)` to the circle `2x^2+2y^2-25` will be perpendicular to each other if `alpha` equals 5 (b) `-4` (c) 4 (d) `-5`

Options

A. 5
B. `-4`
C. 4
D. `-5`

Explanation

To solve the problem, we need to find the value of \( \alpha \) such that the tangents drawn from the point \( (\alpha, 3) \) to the circle \( 2x^2 + 2y^2 - 25 = 0 \) are perpendicular to each other.

### Step 1: Rewrite the equation of the circle
The given equation of the circle is:
\[ 2x^2 + 2y^2 - 25 = 0 \]
Dividing the entire equation by 2, we get:
\[ x^2 + y^2 = \frac{25}{2} \]

### Step 2: Identify the radius of the circle
From the equation \( x^2 + y^2 = \frac{25}{2} \), we can see that the radius \( r \) of the circle is:
\[ r = \sqrt{\frac{25}{2}} = \frac{5}{\sqrt{2}} \]

### Step 3: Find the equation of the director circle
The director circle of a circle \( x^2 + y^2 = r^2 \) is given by:
\[ x^2 + y^2 = 2r^2 \]
Substituting \( r^2 = \frac{25}{2} \):
\[ x^2 + y^2 = 2 \times \frac{25}{2} = 25 \]

### Step 4: Set up the condition for perpendicular tangents
For the tangents drawn from the point \( (\alpha, 3) \) to be perpendicular, the point must lie on the director circle:
\[ \alpha^2 + 3^2 = 25 \]

### Step 5: Solve for \( \alpha \)
Substituting \( 3^2 = 9 \):
\[ \alpha^2 + 9 = 25 \]
Subtracting 9 from both sides:
\[ \alpha^2 = 25 - 9 = 16 \]
Taking the square root of both sides gives:
\[ \alpha = \pm 4 \]

### Final Answer
Thus, the possible values of \( \alpha \) are \( 4 \) and \( -4 \).