The locus of the point, the chord of contact of which wrt the circle `x^(2)+y^(2)=a^(2)` subtends a right angle at the centre of the circle is

Options

A. `x^(2)+y^(2)=a^(2)//2`
B. `x^(2)+y^()=a^(2)//3`
C. `x^(2)+y^(2)=2a^(2)`
D. `x^(2)+y^(2)=3a^(2)`

Answer

To solve the problem, we need to find the locus of the point from which the chord of contact with respect to the circle \(x^2 + y^2 = a^2\) subtends a right angle at the center of the circle.

### Step-by-Step Solution:

1. **Understanding the Circle**:
The given circle is represented by the equation \(x^2 + y^2 = a^2\). The center of the circle is at the origin (0, 0) and the radius is \(a\).

2. **Chord of Contact**:
For a point \(P(h, k)\) outside the circle, the chord of contact can be expressed using the formula:
\[ hx + ky = a^2 \]
This line represents the chord of contact from point \(P\) to the circle.

3. **Condition for Right Angle**:
The problem states that this chord of contact subtends a right angle at the center of the circle. This means that the angles formed by the tangents at the points where the chord touches the circle are \(90^\circ\).

4. **Geometric Interpretation**:
If we denote the points where the chord of contact intersects the circle as \(A\) and \(B\), we can observe that triangles \(APC\) and \(BPC\) (where \(C\) is the center of the circle) are right triangles.

5. **Using Triangle Properties**:
Since \(APC\) and \(BPC\) are congruent (by RHS criterion), we have:
\[ \angle APC = \angle BPC \]
Given that the total angle \(APB\) is \(90^\circ\), we can conclude:
\[ \angle APC = \angle BPC = 45^\circ \]

6. **Applying Sine Rule**:
In triangle \(APC\), we can use the sine of \(45^\circ\):
\[ \sin 45^\circ = \frac{AC}{PC} \]
Since \(\sin 45^\circ = \frac{1}{\sqrt{2}}\), we have:
\[ \frac{1}{\sqrt{2}} = \frac{AC}{PC} \]

7. **Finding Lengths**:
The length \(AC\) is equal to the radius \(a\), and the length \(PC\) can be calculated using the distance formula:
\[ PC = \sqrt{h^2 + k^2} \]
Thus, substituting the values, we get:
\[ \frac{1}{\sqrt{2}} = \frac{a}{\sqrt{h^2 + k^2}} \]

8. **Squaring Both Sides**:
Squaring both sides gives:
\[ \frac{1}{2} = \frac{a^2}{h^2 + k^2} \]
Rearranging this leads to:
\[ h^2 + k^2 = 2a^2 \]

9. **Substituting Variables**:
Finally, replace \(h\) with \(x\) and \(k\) with \(y\) to express the locus:
\[ x^2 + y^2 = 2a^2 \]

### Conclusion:
The locus of the point \(P\) is given by the equation:
\[ x^2 + y^2 = 2a^2 \]