Show that the locus of P where the tangents drawn from P to the circle `x^(2)+y^(2)=a^(2)` include an angle `alpha` is `x^(2)+y^(2)=a^(2)cosec^(2)(alpha)/2`

Answer

To show that the locus of point P, where the tangents drawn from P to the circle \(x^2 + y^2 = a^2\) include an angle \(\alpha\), is given by the equation \(x^2 + y^2 = a^2 \csc^2(\alpha/2)\), we can follow these steps:

### Step 1: Understand the Geometry
The circle given is centered at the origin (0, 0) with radius \(a\). The point \(P(h, k)\) is outside the circle from which tangents are drawn.

### Step 2: Use the Tangent Length Formula
The length of the tangent \(L\) from a point \(P(h, k)\) to the circle \(x^2 + y^2 = a^2\) is given by:
\[ L = \sqrt{h^2 + k^2 - a^2} \]

### Step 3: Relate the Angle to Tangent Length
The angle \(\alpha\) between the two tangents drawn from point \(P\) can be related to the lengths of the tangents. The formula for the angle between two tangents from a point to a circle is:
\[ \tan\left(\frac{\alpha}{2}\right) = \frac{L}{d} \]
where \(d\) is the distance from the point \(P\) to the center of the circle, which is \(d = \sqrt{h^2 + k^2}\).

### Step 4: Substitute the Length of the Tangent
Substituting the expression for \(L\):
\[ \tan\left(\frac{\alpha}{2}\right) = \frac{\sqrt{h^2 + k^2 - a^2}}{\sqrt{h^2 + k^2}} \]

### Step 5: Square Both Sides
Squaring both sides gives:
\[ \tan^2\left(\frac{\alpha}{2}\right) = \frac{h^2 + k^2 - a^2}{h^2 + k^2} \]

### Step 6: Use the Identity for Tangent
Using the identity \(\tan^2\left(\frac{\alpha}{2}\right) = \frac{1 - \cos\alpha}{1 + \cos\alpha}\) and substituting it into the equation gives:
\[ \frac{1 - \cos\alpha}{1 + \cos\alpha} = \frac{h^2 + k^2 - a^2}{h^2 + k^2} \]

### Step 7: Cross Multiply and Rearrange
Cross multiplying gives:
\[ (1 - \cos\alpha)(h^2 + k^2) = (1 + \cos\alpha)(h^2 + k^2 - a^2) \]
Expanding and rearranging leads to:
\[ h^2 + k^2 - a^2\cos\alpha = a^2 \]

### Step 8: Final Equation
Rearranging gives:
\[ h^2 + k^2 = a^2 \frac{1 + \cos\alpha}{1 - \cos\alpha} = a^2 \csc^2\left(\frac{\alpha}{2}\right) \]

This shows that the locus of point \(P\) is:
\[ x^2 + y^2 = a^2 \csc^2\left(\frac{\alpha}{2}\right) \]