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 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 \(x^2 + y^2 = b^2\) that touches the circle \(x^2 + y^2 - 2by = 0\).

### Step 1: Identify the circles and their properties
The first circle is given by:
\[ x^2 + y^2 = b^2 \]
This circle has its center at \((0, 0)\) and radius \(b\).

The second circle is given by:
\[ x^2 + (y - b)^2 = b^2 \]
This can be rewritten as:
\[ x^2 + y^2 - 2by + b^2 = b^2 \implies x^2 + y^2 - 2by = 0 \]
This circle has its center at \((0, b)\) and radius \(b\).

### Step 2: Equation of the chord of contact
Let \(P(h, k)\) be the point from which tangents are drawn to the first circle. The equation of the chord of contact from point \(P\) to the first circle is given by:
\[ xh + yk = b^2 \]
Rearranging gives:
\[ xh + yk - b^2 = 0 \quad \text{(Equation 1)} \]

### Step 3: Perpendicular distance from the center of the second circle
The center of the second circle is \((0, b)\). The perpendicular distance \(d\) from this center to the line given by Equation 1 must equal the radius \(b\) of the second circle. The formula for the perpendicular distance from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our line \(xh + yk - b^2 = 0\), we have \(A = h\), \(B = k\), and \(C = -b^2\). Thus, the distance from \((0, b)\) is:
\[ d = \frac{|h(0) + k(b) - b^2|}{\sqrt{h^2 + k^2}} = \frac{|kb - b^2|}{\sqrt{h^2 + k^2}} \]
Setting this equal to the radius \(b\):
\[ \frac{|kb - b^2|}{\sqrt{h^2 + k^2}} = b \]

### Step 4: Simplifying the equation
Cross-multiplying gives:
\[ |kb - b^2| = b\sqrt{h^2 + k^2} \]
This leads to two cases:
1. \(kb - b^2 = b\sqrt{h^2 + k^2}\)
2. \(kb - b^2 = -b\sqrt{h^2 + k^2}\)

### Step 5: Solving for \(k\)
Let’s consider the first case:
\[ kb - b^2 = b\sqrt{h^2 + k^2} \]
Rearranging gives:
\[ kb = b^2 + b\sqrt{h^2 + k^2} \]
Dividing through by \(b\) (assuming \(b \neq 0\)):
\[ k = b + \sqrt{h^2 + k^2} \]
Squaring both sides:
\[ k^2 = (b + \sqrt{h^2 + k^2})^2 \]
Expanding gives:
\[ k^2 = b^2 + 2b\sqrt{h^2 + k^2} + h^2 + k^2 \]
Cancelling \(k^2\) from both sides:
\[ 0 = b^2 + 2b\sqrt{h^2 + k^2} + h^2 \]
This leads to:
\[ 2b\sqrt{h^2 + k^2} = - (b^2 + h^2) \]
This equation will help us find the locus.

### Step 6: Locus of the point
We can express \(k\) in terms of \(h\) and find the locus:
\[ k = \frac{b^2 - h^2}{2b} \]
Thus, the locus of the point \((h, k)\) is given by:
\[ y = \frac{b^2 - x^2}{2b} \]

### Conclusion
The locus of the point of intersection of the tangents at the extremities of the chord of the first circle that touches the second circle is:
\[ y = \frac{b^2 - x^2}{2b} \]