Question:

Step 1: Identify parameters of the given circles. The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. For Circle 1: $x^2 + y^2 + 4x - 6y + 9 = 0$ Comparing with the general form, we have: $2g_1 = 4 \Rightarrow g_1 = 2$ $2f_1 = -6 \Rightarrow f_1 = -3$ $c_1 = 9$ For Circle 2: $x^2 + y^2 - 5x + 4y - 2 = 0$ Comparing with the general form, we have: $2g_2 = -5 \Rightarrow g_2 = -\frac{5}{2}$ $2f_2 = 4 \Rightarrow f_2 = 2$ $c_2 = -2$ Step 2: Apply the orthogonality condition. Let the center of a circle that cuts the given circles orthogonally be $(h, k)$. The equation of such a circle can be written as $x^2 + y^2 - 2hx - 2ky + (h^2+k^2-R^2) = 0$, where $R$ is its radius. For this circle, we have $g = -h$, $f = -k$, and $c_{new} = h^2+k^2-R^2$. The condition for two circles $x^2 + y^2 + 2g_Ax + 2f_Ay + c_A = 0$ and $x^2 + y^2 + 2g_Bx + 2f_By + c_B = 0$ to cut orthogonally is $2g_Ag_B + 2f_Af_B = c_A + c_B$. Applying this condition for the circle with center $(h, k)$ and Circle 1: $$2g_1(-h) + 2f_1(-k) = c_1 + (h^2+k^2-R^2)$$ $$2(2)(-h) + 2(-3)(-k) = 9 + h^2+k^2-R^2$$ $$-4h + 6k = 9 + h^2+k^2-R^2 \quad \text{(Equation 1)}$$ Applying this condition for the circle with center $(h, k)$ and Circle 2: $$2g_2(-h) + 2f_2(-k) = c_2 + (h^2+k^2-R^2)$$ $$2\left(-\frac{5}{2}\right)(-h) + 2(2)(-k) = -2 + h^2+k^2-R^2$$ $$5h - 4k = -2 + h^2+k^2-R^2 \quad \text{(Equation 2)}$$ Step 3: Determine the locus of the center. Subtract Equation 2 from Equation 1 to eliminate the term $h^2+k^2-R^2$: $$(-4h + 6k) - (5h - 4k) = (9 + h^2+k^2-R^2) - (-2 + h^2+k^2-R^2)$$ $$-4h + 6k - 5h + 4k = 9 + 2$$ $$-9h + 10k = 11$$ $$9h - 10k = -11$$ Step 4: Express the locus equation. Replacing $h$ with $x$ and $k$ with $y$, the locus of the centers of the circles is: $$9x - 10y = -11$$ $$9x - 10y + 11 = 0$$