The locus of points of intersection of the tangents to `x^(2)+y^(2)=a^(2)` at the extremeties of a chord of circle `x^(2)+y^(2)=a^(2)` which touches the circle `x^(2)+y^(2)-2ax=0` is/are :

Options

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

Answer

To solve the problem, we need to find the locus of points of intersection of the tangents to the circle \(x^2 + y^2 = a^2\) at the extremities of a chord that touches another circle \(x^2 + y^2 - 2ax = 0\).

### Step 1: Identify the circles
1. The first circle is given by the equation \(x^2 + y^2 = a^2\).
- Center: \((0, 0)\)
- Radius: \(a\)

2. The second circle can be rewritten as:
\[ x^2 + y^2 - 2ax = 0 \implies (x - a)^2 + y^2 = a^2 \]
- Center: \((a, 0)\)
- Radius: \(a\)

### Step 2: Equation of the tangent to the first circle
The equation of the tangent to the circle \(x^2 + y^2 = a^2\) at a point \((x_1, y_1)\) on the circle is given by:
\[ xx_1 + yy_1 = a^2 \]
For the locus, we will replace \((x_1, y_1)\) with \((h, k)\), where \((h, k)\) is the point of intersection.

### Step 3: Write the equation of the tangent in terms of \(h\) and \(k\)
The equation becomes:
\[ hx + ky = a^2 \tag{1} \]

### Step 4: Find the distance from the center of the second circle to the line (1)
The distance \(d\) from the center \((a, 0)\) of the second circle to the line \(hx + ky = a^2\) is given by:
\[ d = \frac{|ha + k \cdot 0 - a^2|}{\sqrt{h^2 + k^2}} = \frac{|ha - a^2|}{\sqrt{h^2 + k^2}} \]
Since the chord touches the second circle, this distance must equal the radius \(a\):
\[ \frac{|ha - a^2|}{\sqrt{h^2 + k^2}} = a \]

### Step 5: Simplify the equation
Squaring both sides gives:
\[ (ha - a^2)^2 = a^2(h^2 + k^2) \]
Expanding both sides:
\[ h^2a^2 - 2ha^2 + a^4 = a^2h^2 + a^2k^2 \]
Rearranging terms:
\[ h^2a^2 - a^2h^2 - 2ha^2 + a^4 - a^2k^2 = 0 \]
This simplifies to:
\[ -2ha^2 + a^4 - a^2k^2 = 0 \]
Dividing through by \(a^2\) (assuming \(a \neq 0\)):
\[ -2h + a^2 - k^2 = 0 \]
Rearranging gives:
\[ k^2 = a^2 - 2h \tag{2} \]

### Step 6: Final equations
From equation (2), we can express the locus:
1. \(k^2 = a^2 - 2h\) can be rewritten as:
\[ y^2 = a^2 - 2x \]
2. The other form can be derived from the original tangent equation, leading to:
\[ x^2 + y^2 = (x - a)^2 \]
which simplifies to:
\[ x^2 + y^2 = x^2 - 2ax + a^2 \implies y^2 = a^2 - 2ax \]

### Conclusion
The locus of points of intersection of the tangents is given by:
1. \(y^2 = a^2 - 2x\)
2. \(x^2 + y^2 = (x - a)^2\)