To find the locus of the point whose chord of contact with respect to the circle \( x^2 + y^2 = a^2 \) makes an angle \( 2\alpha \) at the center of the circle, we can follow these steps:
### Step 1: Understand the Circle and Chord of Contact
The given circle has the equation \( x^2 + y^2 = a^2 \), where the center is at the origin \( (0, 0) \) and the radius is \( r = a \). A point \( P(h, k) \) outside the circle can have two tangents drawn to the circle, which will touch the circle at points \( A \) and \( B \). The line segment \( AB \) is the chord of contact.
**Hint:** Visualize the circle and the point outside it. Draw the tangents from the point to the circle.
### Step 2: Equation of the Chord of Contact
The equation of the chord of contact from the point \( P(h, k) \) to the circle is given by:
\[
hx + ky = a^2
\]
**Hint:** Remember that the chord of contact relates the coordinates of the external point to the circle's radius.
### Step 3: Perpendicular Distance from Center to Chord
The perpendicular distance \( OM \) from the center \( O(0, 0) \) to the chord \( AB \) can be calculated using the formula for the distance from a point to a line. The formula is:
\[
d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}
\]
For our chord \( hx + ky - a^2 = 0 \), we have \( A = h \), \( B = k \), and \( C = -a^2 \). Thus, the distance \( OM \) is:
\[
OM = \frac{|h(0) + k(0) - a^2|}{\sqrt{h^2 + k^2}} = \frac{a^2}{\sqrt{h^2 + k^2}}
\]
**Hint:** This distance is crucial as it relates to the angles formed at the center.
### Step 4: Relate the Angle to the Triangle
In triangle \( OAM \), where \( O \) is the center, \( A \) is a point on the circle, and \( M \) is the foot of the perpendicular from \( O \) to the chord \( AB \), we can use the cosine of angle \( \alpha \):
\[
\cos(\alpha) = \frac{OM}{OA}
\]
Since \( OA = a \) (the radius of the circle), we have:
\[
\cos(\alpha) = \frac{OM}{a}
\]
**Hint:** Use trigonometric identities to relate the angle to the sides of the triangle.
### Step 5: Substitute the Distance
Substituting \( OM \):
\[
\cos(\alpha) = \frac{\frac{a^2}{\sqrt{h^2 + k^2}}}{a}
\]
This simplifies to:
\[
\cos(\alpha) = \frac{a}{\sqrt{h^2 + k^2}}
\]
### Step 6: Square Both Sides
Squaring both sides gives:
\[
\cos^2(\alpha) = \frac{a^2}{h^2 + k^2}
\]
Rearranging this, we find:
\[
h^2 + k^2 = \frac{a^2}{\cos^2(\alpha)}
\]
**Hint:** This equation represents the relationship between the coordinates \( (h, k) \) and the angle \( \alpha \).
### Step 7: Final Locus Equation
Letting \( h = x \) and \( k = y \), we rewrite the equation as:
\[
x^2 + y^2 = a^2 \sec^2(\alpha)
\]
**Hint:** Recognize that \( \sec^2(\alpha) = \frac{1}{\cos^2(\alpha)} \).
### Conclusion
Thus, the locus of the point \( P(h, k) \) is given by:
\[
x^2 + y^2 = a^2 \sec^2(\alpha)
\]
This is the required locus of the point whose chord of contact with respect to the circle makes an angle \( 2\alpha \) at the center.
