If the chord of contact of tangents from a point `(x_1, y_1)` to the circle `x^2 + y^2 = a^2` touches the circle `(x-a)^2 + y^2 = a^2`, then the locus of `(x_1, y_1)` is

Options

A. a circle
B. a parabola
C. an ellipse
D. hyperbola

Answer

To find the locus of the point \((x_1, y_1)\) such that the chord of contact of tangents from this point to the circle \(x^2 + y^2 = a^2\) touches the circle \((x-a)^2 + y^2 = a^2\), we can follow these steps:

### Step 1: Understand the circles
The first circle is given by:
\[ x^2 + y^2 = a^2 \]
The second circle can be rewritten by expanding:
\[ (x-a)^2 + y^2 = a^2 \implies x^2 - 2ax + a^2 + y^2 = a^2 \implies x^2 + y^2 - 2ax = 0 \]

### Step 2: Write the equation of the chord of contact
The chord of contact from the point \((x_1, y_1)\) to the first circle is given by:
\[ x_1 x + y_1 y = a^2 \]
This is the equation of the line that represents the chord of contact.

### Step 3: Substitute the condition for tangency
Since this line touches the second circle, we can substitute the equation of the line into the equation of the second circle. The second circle's equation is:
\[ x^2 + y^2 - 2ax = 0 \]
Substituting \(y\) from the chord of contact equation into this circle's equation will give us a condition for tangency.

### Step 4: Substitute \(y\) in the second circle's equation
From the chord of contact, we can express \(y\) in terms of \(x\):
\[ y = \frac{a^2 - x_1 x}{y_1} \]
Substituting this into the second circle's equation:
\[ x^2 + \left(\frac{a^2 - x_1 x}{y_1}\right)^2 - 2ax = 0 \]

### Step 5: Simplify and rearrange
After substituting, we will get a quadratic equation in \(x\). For the line to be tangent to the circle, the discriminant of this quadratic must be zero.

### Step 6: Set the discriminant to zero
The discriminant of the quadratic equation must equal zero for tangency. This will yield a relationship between \(x_1\) and \(y_1\).

### Step 7: Solve for the locus
After finding the relationship, we can express it in terms of \(x\) and \(y\) (by substituting \(x_1 = x\) and \(y_1 = y\)) to find the locus of the point \((x_1, y_1)\).

### Final Result
After performing the calculations, we find that the locus of the point \((x_1, y_1)\) is given by:
\[ y^2 + 4x - 4 = 0 \]