Let $S_1 = x^2 + y^2 - kx = 0$ and $S_2 = x^2 - y^2 - cx = 0$, then
$S_1$ and $S_2$ intersect at an angle of $\pi/4$
$S_1$ and $S_2$ intersect orthogonally
abscissa of the point of intersection of $S_1$ and $S_2$ is A.M. of $c$ and $k$
point of intersection of $S_1$ and $S_2$ is origin
Step-by-Step Solution
Key Concept: Find intersection points of $S_1: x^2 + y^2 - kx = 0$ (circle) and $S_2: x^2 - y^2 - cx = 0$ (hyperbola) by solving simultaneously. Subtracting $S_2$ from $S_1$ gives $2y^2 - (k-c)x = 0$, which shows the abscissa satisfies $x = rac{2y^2}{k-c}$. When $y = 0$, both curves pass through the origin, and when $y
eq 0$, the x-coordinate relates to the arithmetic mean $(k+c)/2$.
From $x^2 + y^2 = kx$, differentiating gives $2x + 2y\frac{dy}{dx} = k$, so $\frac{dx}{dx} = \frac{x^2 + y^2}{x}$. For the orthogonal trajectory, $\frac{dy}{dx} = -\frac{x}{2y - x}$, which simplifies to $\frac{x^2 - y^2}{2xy} = \frac{dx}{dy}$. This gives the orthogonal family after solving.
Correct Answer: 3,4