The locus of the point of intersection of the tangents at the extremities of a chord of the circle `x^(2)+y^(2)=r^(2)` which touches the circle `x^(2)+y^(2)+2rx=0` is

Options

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

Answer

To find the locus of the point of intersection of the tangents at the extremities of a chord of the circle \(x^2 + y^2 = r^2\) which touches the circle \(x^2 + y^2 + 2rx = 0\), we can follow these steps:

### Step 1: Understand the Circle and Tangents
We have two circles:
1. Circle 1: \(x^2 + y^2 = r^2\) (center at (0,0) and radius r)
2. Circle 2: \(x^2 + y^2 + 2rx = 0\) can be rewritten as \((x + r)^2 + y^2 = r^2\) (center at (-r, 0) and radius r)

### Step 2: Equation of the Chord of Contact
For a point \((h, k)\) outside the circle \(x^2 + y^2 = r^2\), the chord of contact is given by:
\[ hx + ky = r^2 \]
This line represents the tangents drawn from the point \((h, k)\) to the first circle.

### Step 3: Distance from the Center of Circle 2
The chord of contact must also touch the second circle. The distance from the center of the second circle (-r, 0) to the line \(hx + ky - r^2 = 0\) should equal the radius r.

Using the distance formula from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\):
\[ \text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
Here, \(A = h\), \(B = k\), and \(C = -r^2\). The center of the second circle is \((-r, 0)\).

### Step 4: Set up the Distance Equation
The distance from the center of the second circle to the chord of contact is:
\[ \frac{|h(-r) + k(0) - r^2|}{\sqrt{h^2 + k^2}} = r \]
This simplifies to:
\[ \frac{|-hr - r^2|}{\sqrt{h^2 + k^2}} = r \]

### Step 5: Solve the Equation
Cross-multiplying gives:
\[ |-hr - r^2| = r\sqrt{h^2 + k^2} \]
This can be split into two cases:
1. \(-hr - r^2 = r\sqrt{h^2 + k^2}\)
2. \(-hr - r^2 = -r\sqrt{h^2 + k^2}\)

### Step 6: Analyze the Cases
For case 1:
\[ hr + r^2 + r\sqrt{h^2 + k^2} = 0 \]
For case 2:
\[ hr + r^2 - r\sqrt{h^2 + k^2} = 0 \]

### Step 7: Rearranging and Squaring
From case 1, we can rearrange to find:
\[ \sqrt{h^2 + k^2} = -\frac{hr + r^2}{r} \]
Squaring both sides leads to:
\[ h^2 + k^2 = \left(\frac{hr + r^2}{r}\right)^2 \]

### Step 8: Final Equation
After simplifying, we arrive at:
\[ k^2 = 2hr + r^2 \]
Substituting \(h = x\) and \(k = y\):
\[ y^2 = 2rx + r^2 \]

### Conclusion
Thus, the locus of the point of intersection of the tangents at the extremities of the chord is given by:
\[ y^2 = 2rx + r^2 \]