The points of intersection of the line ax + by = 0, (a ≠ b) and the circle x² + y² - 2x = 0 are A(a, 0) and B(1, β). The image of the circle with AB as a diameter in the line x + y + 2 = 0 is

Step-by-step solution

1. 1

   Find a and β

   A(a,0) on circle • a² + 0² - 2a = 0 • a(a-2) = 0 → a=2 (a≠0)

   $$a^2 - 2a = 0$$
2. 2

   Find β

   B(1,β) on circle • 1² + β² - 2(1) = 0 • β² - 1 = 0 → β = ±1

   $$1 + \beta^2 - 2 = 0$$
3. 3

   Determine b and β

   A(a,0) on x²+y²-2x=0 • a²-2a=0 → a=0 or a=2 • B(1,β) on x²+y²-2x=0
4. 4

   Circle with AB as diameter

   A(2,0), B(1,1) • Center C = ((2+1)/2, (0+1)/2) = (3/2, 1/2) • Radius² = ((2-1)² + (0-1)²)/4 = (1+1)/4 = 2/4 = 1/2

   $$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$$
5. 5

   Image of the circle

   Center C(3/2, 1/2) • Line L: x+y+2=0 • Image C'(x',y')

   $$\frac{x'-x_1}{A} = \frac{y'-y_1}{B} = \frac{-2(Ax_1+By_1+C)}{A^2+B^2}$$