The locus of the point of intersection of the tangents at the extermities of a chord of the circle `x^2+y^2=b^2` which touches the circle `x^2+y^2-2by=0` passes through the point

Options

A. `(0,b/2)`
B. (0,b)
C. (b,0)
D. `(b/2,0)`

Answer

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

### Step 1: Understand the circles
The first circle is given by the equation \(x^2 + y^2 = b^2\), which is a circle centered at the origin with radius \(b\). The second circle is given by \(x^2 + (y - b)^2 = b^2\), which is a circle centered at \((0, b)\) with radius \(b\).

### Step 2: Identify the tangents
For a chord of the first circle, let the points at the extremities be \(A(x_1, y_1)\) and \(B(x_2, y_2)\). The tangents at these points can be expressed using the point of tangency and the slope of the tangent line.

### Step 3: Equation of the tangent
The equation of the tangent to the circle \(x^2 + y^2 = b^2\) at a point \((x_1, y_1)\) is given by:
\[ xx_1 + yy_1 = b^2 \]
Similarly, for the point \((x_2, y_2)\), the tangent equation is:
\[ xx_2 + yy_2 = b^2 \]

### Step 4: Find the point of intersection
To find the point of intersection of the two tangents, we can solve the two equations simultaneously:
1. \(xx_1 + yy_1 = b^2\)
2. \(xx_2 + yy_2 = b^2\)

By subtracting these two equations, we can eliminate \(b^2\) and find a relationship between \(x\) and \(y\).

### Step 5: Use the condition of tangency
Since the chord touches the second circle, the distance from the center of the second circle to the line formed by the tangents must equal the radius of the second circle. This gives us a condition that relates the coordinates of the intersection point to the radius \(b\).

### Step 6: Derive the locus equation
Using the relationships derived from the previous steps, we can express the locus of the point of intersection in terms of \(x\) and \(y\). After simplification, we will arrive at an equation of the form:
\[ x^2 = -2by + b^2 \]
This represents a parabola.

### Step 7: Identify points through which the locus passes
To find specific points through which the locus passes, we can substitute values into the derived equation.

1. Check the point \((0, \frac{b}{2})\):
\[ 0^2 = -2b(\frac{b}{2}) + b^2 \implies 0 = -b^2 + b^2 \implies 0 = 0 \quad \text{(True)} \]

2. Check the point \((b, 0)\):
\[ b^2 = -2b(0) + b^2 \implies b^2 = b^2 \quad \text{(True)} \]

3. Check the point \((\frac{b}{2}, 0)\):
\[ \left(\frac{b}{2}\right)^2 = -2b(0) + b^2 \implies \frac{b^2}{4} = b^2 \quad \text{(False)} \]

### Conclusion
The locus of the point of intersection of the tangents at the extremities of the chord passes through the points \((0, \frac{b}{2})\) and \((b, 0)\).