Find the points of intersection of the circles $x^2 + y^2 = 130$ and $x^2 + y^2 - 10x - 2y - 26 = 0$ and find the tangents at their common points.
Equations of given circles:
$x^2 + y^2 = 130$ $\;\;\; \cdots \; (1)$
$x^2 + y^2 - 10x - 2y - 26 = 0$ $\;\;\; \cdots \; (2)$
In view of equation $(1)$, equation $(2)$ can be written as
$130 - 10x - 2y - 26 = 0$
i.e. $\;$ $10x + 2y = 104$
$\implies$ $y = 52 - 5x$ $\;\;\; \cdots \; (3)$
In view of equation $(3)$ equation $(1)$ becomes
$x^2 + \left(52 - 5x\right)^2 = 130$
i.e. $\;$ $x^2 + 2704 - 520 x + 25x^2 = 130$
i.e. $\;$ $26x^2 - 520x + 2574 = 0$
i.e. $\;$ $x^2 - 20 x + 99 = 0$
$\implies$ $x = 11$ $\;$ or $\;$ $x = 9$
We have from equation $(3)$,
when $\;$ $x = 11$, $\;$ $y = 52 - 5 \times 11 = -3$
when $\;$ $x = 9$, $\;$ $y = 52 - 5 \times 9 = 7$
$\therefore \;$ The points of intersections of the two given circles are
$\left(11, -3\right)$ $\;$ and $\;$ $\left(9, 7\right)$
Equation of tangent to the circle $x^2 + y^2 = r^2$ at any given point $\left(x_1, y_1\right)$ is
$xx_1 + yy_1 = r^2$
$\therefore \;$ Equation of tangent to circle given by equation $(1)$ at $\left(11, -3\right)$ is
$11x - 3y = 130$
Equation of tangent to circle given by equation $(1)$ at $\left(9, 7\right)$ is
$9x + 7y = 130$