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 solve the problem, we need to find the locus of the point of intersection of the tangents at the extremities of a chord of the circle given by the equation \(x^2 + y^2 = r^2\), which touches another circle given by the equation \(x^2 + y^2 + 2rx = 0\).

### Step 1: Understand the circles
The first circle is centered at the origin \((0, 0)\) with radius \(r\). The second circle can be rewritten as:
\[ x^2 + y^2 = -2rx \]
This represents a circle with center at \((-r, 0)\) and radius \(r\).

### Step 2: Find the chord of contact
For a point \(P(h, k)\) outside the circle \(x^2 + y^2 = r^2\), the equation of the chord of contact is given by:
\[ hx + ky = r^2 \]
Rearranging gives us:
\[ hx + ky - r^2 = 0 \]

### Step 3: Condition for tangency
The chord of contact must touch the second circle. The condition for tangency of a line \(Ax + By + C = 0\) to a circle centered at \((g, f)\) with radius \(R\) is given by:
\[ \frac{|Ag + Bf + C|}{\sqrt{A^2 + B^2}} = R \]
For our case, \(A = h\), \(B = k\), \(C = -r^2\), \(g = -r\), \(f = 0\), and \(R = r\).

### Step 4: Apply the tangency condition
Substituting into the tangency condition:
\[ \frac{|h(-r) + k(0) - r^2|}{\sqrt{h^2 + k^2}} = r \]
This simplifies to:
\[ \frac{|-hr - r^2|}{\sqrt{h^2 + k^2}} = r \]
Multiplying both sides by \(\sqrt{h^2 + k^2}\) gives:
\[ |-hr - r^2| = r\sqrt{h^2 + k^2} \]

### Step 5: Square both sides
Squaring both sides results in:
\[ (-hr - r^2)^2 = r^2(h^2 + k^2) \]
Expanding the left-hand side:
\[ (h^2r^2 + 2hr^3 + r^4) = r^2(h^2 + k^2) \]
Rearranging gives:
\[ h^2r^2 + 2hr^3 + r^4 - r^2h^2 - r^2k^2 = 0 \]
This simplifies to:
\[ 2hr^3 + r^4 - r^2k^2 = 0 \]

### Step 6: Solve for \(k^2\)
Rearranging yields:
\[ r^2k^2 = r^4 + 2hr^3 \]
Dividing by \(r^2\):
\[ k^2 = r^2 + 2h\frac{r^3}{r^2} \]
This simplifies to:
\[ k^2 = r^2 + 2hr \]

### Step 7: Express the locus
Rearranging gives:
\[ k^2 = 2hr + r^2 \]
This is the equation of a parabola in the form:
\[ y^2 = 2rx + r^2 \]
Replacing \(h\) and \(k\) with \(x\) and \(y\) respectively, we get:
\[ y^2 = 2rx + r^2 \]

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