A line L intersects the circle S: x^2 + y^2 - 6x - 2y + 6 = 0 at points A and B . If the image of the circle drawn with AB as its diameter in the line x + y = 0 is the circle x^2 + y^2 + 4x + 4y + 6 = 0 , then the equation of the line L is :

Step-by-step solution

The equation of the image circle is x^2 + y^2 + 4x + 4y + 6 = 0 . The center of this image circle is C'(-2, -2) . Let the center of the circle with AB as diameter be C(h, k) . Since C' is the reflection of C in the line x + y = 0 , we can find (h, k) using the reflection formula: -2 - h 1 = -2 - k 1 = -2(h + k) 1^2 + 1^2 This gives h = 2 and k = 2 . Thus, the center of the circle with diameter AB is C(2, 2) . Since AB is the diameter, C(2, 2) is the midpoint of the chord AB of the original circle S: x^2 + y^2 - 6x