A circle passes through the origin whose center lies on x^2 + y^2 - Hx - 6y + 10 = 0 and it cuts orthogonally.

Steps to Solve

1. Identify the center of the circle Let the center of the circle be denoted as $(h, k)$. Since the center lies on the curve defined by $x^2 + y^2 - Hx - 6y + 10 = 0$, we substitute $h$ and $k$ into that equation: $$ h^2 + k^2 - Hh - 6k + 10 = 0 $$

2. Use the circle's equation The equation of a circle passing through the origin and having a center at $(h, k)$ is given by: $$ (x - h)^2 + (y - k)^2 = r^2 $$ Since it passes through the origin $(0, 0)$, substituting these values gives: $$ h^2 + k^2 = r^2 $$

3. Orthogonality condition For the circle to intersect orthogonally with the given curve, the condition is that: $$ 2h \cdot H + 2k \cdot 6 = r^2 $$ Substituting $r^2$ from the previous step, we have: $$ 2hH + 12k = h^2 + k^2 $$

4. Formulate equations Now we have the system of equations:

5. $$ h^2 + k^2 - Hh - 6k + 10 = 0 $$

6. $$ 2hH + 12k = h^2 + k^2 $$

7. Solve the system From the first equation, express $h^2 + k^2$ in terms of $h$ and $k$: $$ h^2 + k^2 = Hh + 6k - 10 $$ Substituting into the second equation results in: $$ 2hH + 12k = Hh + 6k - 10 $$

8. Rearranging Rearranging gives: $$ Hh + 6k - 2hH - 12k + 10 = 0 $$ Simplifying leads to: $$ (1 - 2H)h + (6 - 12)k + 10 = 0 $$ Which gives the relation between $h$ and $k$.

9. Final form of circle equation Using the relations deduced, substitute back into the circle equation to express it in standard form.