Find the radical axis of the circles $x^2 + y^2 - 4x + 6y - 10 = 0$ and $x^2 + y^2 + 2x - 6y + 2 = 0$ and show that the circles intersect in real points.
Given circles: $\;$ $x^2 + y^2 - 4x + 6y - 10 = 0$ $\;\;\; \cdots \; (1)$
$x^2 + y^2 + 2x - 6y + 2 = 0$ $\;\;\; \cdots \; (2)$
Comparing equation $(1)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives
$g = -2, \; f = 3, \; c = -10$
Comparing equation $(2)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_1 x + 2f_1 y + c_1 = 0$ $\;$ gives
$g_1 = 1, \; f_1 = -3, \; c_1 = 2$
The radical axis of circles $(1)$ and $(2)$ is
$2 x \left(g - g_1\right) + 2y \left(f - f_1\right) + c - c_1 = 0$
i.e. $\;$ $2x \left(-2-1\right) + 2y \left(3 + 3\right) - 10 - 2 = 0$
i.e. $\;$ $-6x + 12 y - 12 = 0$
i.e. $\;$ $x - 2y + 2 = 0$ $\;\;\; \cdots \; (3)$
Let circles $(1)$ and $(2)$ have a common point $\left(x_1, y_1\right)$
Then, $\;$ $x_1^2 + y_1^2 - 4 x_1 + 6 y_1 - 10 = 0$
and $\;$ $x_1^2 + y_1^2 + 2x_1 - 6 y_1 + 2 = 0$
$\therefore \;$ By subtraction, $\;$ $- 6x_1 + 12 y_1 - 12 = 0$
i.e. $\;$ $x_1 - 2 y_1 + 2 = 0$
$\therefore \;$ The equation of the common chord of the two circles is
$x - 2y + 2 = 0$ $\;\;\; \cdots \; (4)$
which is the same as equation $(3)$.
i.e. $\;$ Radical axis is the common chord of the circles.
$\implies$ The two circles intersect in real points.