A circle passes through the origin whose centre lies on αx² + y² - Hx - 6y + 10 = 0 and it cuts the curve orthogonally.

Steps to Solve

1. Identify the center of the circle

Let the center of the circle be $(h, k)$. Since the circle passes through the origin, we can express the radius $r$ in terms of $h$ and $k$ as follows: $$ r = \sqrt{h^2 + k^2} $$

2. Equation of the circle

The equation of the circle can be written as: $$ (x - h)^2 + (y - k)^2 = r^2 $$ Substituting for $r$, we have: $$ (x - h)^2 + (y - k)^2 = h^2 + k^2 $$

3. Substituting the center's constraint

Since the center $(h, k)$ lies on the curve given by the equation $\alpha x^2 + y^2 - Hx - 6y + 10 = 0$, we can substitute $(h, k)$ into this equation: $$ \alpha h^2 + k^2 - Hh - 6k + 10 = 0 $$

4. Orthogonality condition

For the circle to intersect orthogonally with the curve, the following relationship must hold: $$ 2(x_1 - h)(x_2 - h) + 2(y_1 - k)(y_2 - k) = r^2 $$ We need to work with this condition to find values of $h$ and $k$ that satisfy both the circle’s and the curve’s equations.

5. Simplifying the equations

Expand and simplify the equations derived from the circle and orthogonality condition:

• From the equation of the circle: $$ x^2 - 2hx + y^2 - 2ky + (h^2 + k^2 - r^2) = 0 $$
• Using the orthogonality condition, you can substitute various points $(x_1, y_1)$ and $(x_2, y_2)$ based on the curve's equation and relate them back to $(h, k)$.

6. Solving the system

Solve the system of equations involving $(h, k)$ both from the curve equation and the orthogonality condition to find specific values for $\alpha, H, k$.