Question

The locus of the poles of the tangents to the circle \( x^2 + y^2 - 2x + 2y - 2 = 0 \) with respect to the circle \( x^2 + y^2 = 4 \) is

• A. \( 3x^2 + 2xy + 3y^2 + 8x - 8y - 16 = 0 \)
• B. \( x^2 - 2xy + y^2 - 4x + 4y + 8 = 0 \)
• C. \( 3x^2 - 2xy - 3y^2 + 4x - 4y + 16 = 0 \)
• D. \( x^2 + y^2 - 4x + 4y - 8 = 0 \)

Hint:

Let the pole be ( (h, k) ). Write the equation of the polar of ( (h, k) ) with respect to the second circle. Use the condition that this polar is a tangent to the first circle (distance from the center equals the radius). Simplify the resulting equation and replace ( (h, k) ) with ( (x, y) ) to find the locus.

Answer 1

The Correct Option is A

Let the circle be \( S_1: x^2 + y^2 - 2x + 2y - 2 = 0 \) and the circle with respect to which poles are taken be \( S_2: x^2 + y^2 - 4 = 0 \).
The center of \( S_2 \) is \( O(0, 0) \) and its radius is \( R = 2 \).
Let \( P(h, k) \) be the pole of a tangent to \( S_1 \) with respect to \( S_2 \).
The equation of the polar of \( P(h, k) \) with respect to \( S_2 \) is \( hx + ky - 4 = 0 \).
Since this polar is a tangent to \( S_1: (x - 1)^2 + (y + 1)^2 = 1 + 1 + 2 = 4 \), its distance from the center \( (1, -1) \) of \( S_1 \) is equal to the radius \( 2 \).
$$ \frac{|h(1) + k(-1) - 4|}{\sqrt{h^2 + k^2}} = 2 $$ $$ |h - k - 4| = 2\sqrt{h^2 + k^2} $$ Squaring both sides: $$ (h - k - 4)^2 = 4(h^2 + k^2) $$ $$ (h - (k + 4))^2 = 4h^2 + 4k^2 $$ $$ h^2 - 2h(k + 4) + (k + 4)^2 = 4h^2 + 4k^2 $$ $$ h^2 - 2hk - 8h + k^2 + 8k + 16 = 4h^2 + 4k^2 $$ $$ 3h^2 + 3k^2 + 2hk + 8h - 8k - 16 = 0 $$ Replacing \( (h, k) \) with \( (x, y) \), the locus of the pole is: $$ 3x^2 + 3y^2 + 2xy + 8x - 8y - 16 = 0 $$ There seems to be a slight difference with option A.
Let's recheck the algebra.
$$ h^2 - 2hk - 8h + k^2 + 8k + 16 = 4h^2 + 4k^2 $$ $$ 3h^2 + 3k^2 + 2hk + 8h - 8k - 16 = 0 $$ The terms are in a different order in option A.
Final Answer: The final answer is $\boxed{3x^2 + 2xy + 3y^2 + 8x - 8y - 16 = 0}$