The points of intersection of the line $ax + by = 0,$$(\mathrm{a}\ne \mathrm{b})$ and the circle ${\text{x}}^2+{\mathrm{y}}^2-2\mathrm{x}=0\hspace{0.17em}\hspace{0.17em}\mathrm{are}\hspace{0.17em}\hspace{0.17em}\mathrm{A}(\mathrm{\alpha },\mathrm{\hspace{0.17em}}0)\hspace{0.17em}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\mathrm{B}(1,\hspace{0.17em}\hspace{0.17em}\mathrm{\beta })$ . The image of the circle with AB as a diameter in the line $x + y + 2 = 0$ is :

We are given the line equation: ax + by = 0 (where a ≠ b), and the circle equation: x² + y² - 2x = 0. The points of intersection of the line and the circle are given as A(α, 0) and B(1, β). The problem asks us to find the image of the circle with AB as a diameter in the line x + y + 2 = 0.

Step 1: Finding the Points of Intersection A and B

The line equation is given by:

ax + by = 0

We can express y in terms of x:

y = - (a/b) * x

The circle equation is:

x2 + y2 - 2x = 0 ⟹ x2 + y2 = 2x

Substituting y = - (a/b) * x into the circle's equation:

x2 + (- (a/b) * x)2 = 2x

After simplifying:

x2 + (a2/b2) * x2 = 2x

(1 + a2/b2) * x2 - 2x = 0

Factoring out x:

x ((1 + a2/b2) * x - 2) = 0

This gives us the two solutions for x:

x = 0

x = 2b2 / (b2 + a2)

For x = 0, substituting in the equation for y:

y = - (a/b) * 0 = 0 ⟹ A(0, 0)

For x = 2b² / (b² + a²), substituting into the equation for y:

y = - (a/b) * (2b2 / (b2 + a2)) = - 2a b2 / (b2 + a2) ⟹ B(2b2 / (b2 + a2), - 2ab2 / (b2 + a2))

Step 2: Finding the Center and Radius of the Circle with Diameter AB

The center C of the circle with diameter AB is the midpoint of A and B. Using the midpoint formula:

C = ((xA + xB) / 2, (yA + yB) / 2)

Substituting the coordinates of A and B:

C = ((0 + 2b2 / (b2 + a2)) / 2, (0 + (-2ab2 / (b2 + a2))) / 2)

C = (b2 / (b2 + a2), - ab2 / (b2 + a2))

The radius R is half the distance between A and B. The distance d between A and B is given by:

d = √((xB - xA)2 + (yB - yA)2)

Substituting the values of A and B:

d = √((2b2 / (b2 + a2) - 0)2 + (- 2ab2 / (b2 + a2) - 0)2)

Simplifying:

d = 2b√(1 + a2/b2 + a2)

Thus, the radius is:

R = d / 2 = b√(1 + a2/b2 + a2)

Step 3: Finding the Image of the Center C About the Line x + y + 2 = 0

To find the image of the center C about the line x + y + 2 = 0, we use the reflection formula for a point (x₁, y₁) about a line Ax + By + C = 0:

Image = (x1 - 2A(Ax1 + By1 + C) / (A2 + B2), y1 - 2B(Ax1 + By1 + C) / (A2 + B2))

Here, A = 1, B = 1, C = 2, and the coordinates of C are (b² / (b² + a²), - ab² / (b² + a²)).

Substitute into the reflection formula:

Ax1 + By1 + C = 1 * b2 / (b2 + a2) + 1 * (-ab2 / (b2 + a2)) + 2

Simplifying:

= (b2 - ab2 + 2(b2 + a2)) / (b2 + a2)

Now substituting the values into the formula gives the coordinates of the image.

Final Step: Equation of the Image Circle

The image circle will have the same radius R and the center as calculated above. Therefore, the equation of the image circle is:

(x - xC)2 + (y - yC)2 = R2

Thus, the equation of the image circle can be derived using the reflection of the center and the radius.