Show that the point $\left(-9, -1\right)$ lies on the circle $x^2 + y^2 + 4x + 8y - 38 = 0$.
Find the point diametrically opposite to it and find the tangents at these points.
Equation of given circle: $\;$ $x^2 + y^2 + 4x + 8y - 38 = 0$ $\;\;\; \cdots \; (1)$
Given point: $\;$ $P \left(x_1, y_1\right) = \left(-9, -1\right)$
Substituting the given point in the equation of the circle gives
$\left(-9\right)^2 + \left(-1\right)^2 + 4 \times \left(-9\right) + 8 \times \left(-1\right) - 38 = 0$
i.e. $\;$ $81 + 1 - 36 - 8 - 38 = 0$
i.e. $\;$ $0 = 0$ $\;\;$ which is true.
$\implies$ The point $\left(-9, -1\right)$ lies on the given circle.
Comparing the given equation of circle with the standard equation of circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives
$g = 2$, $\;$ $f = 4$, $\;$ $c = -38$
Radius of the given circle $= r = \sqrt{g^2 + f^2 - c}$
i.e. $\;$ $r = \sqrt{2^2 + 4^2 - \left(-38\right)} = \sqrt{4 + 16 + 38} = \sqrt{58}$
$\therefore \;$ Diameter of the circle $= d = 2r = 2 \sqrt{58}$
Let $Q \left(x_2, y_2\right)$ be the point diametrically opposite to $P$.
Then, $PQ$ is the diameter of the circle.
i.e. $\;$ $PQ = \sqrt{\left(x_2 + 9\right)^2 + \left(y_2 + 1\right)^2} = 2\sqrt{58}$
i.e. $\;$ $x_2^2 + 18x_2 + 81 + y_2^2 + 2y_2 + 1 = 232$
i.e. $\;$ $x_2^2 + y_2^2 + 18x_2 + 2y_2 - 150 = 0$ $\;\;\; \cdots \; (2)$
$Q \left(x_2, y_2\right)$ also lies on equation $(1)$.
$\therefore \;$ We have,
$x_2^2 + y_2^2 + 4x_2 + 8y_2 - 38 = 0$ $\;\;\; \cdots \; (3)$
Subtracting equations $(2)$ and $(3)$ gives
$14x_2 - 6y_2 - 112 = 0$
i.e. $\;$ $7x_2 - 3y_2 - 56 = 0$ $\;\;\; \cdots \; (4)$
$\implies$ $x_2 = \dfrac{3y_2 + 56}{7}$
Substituting the value of $x_2$ in equation $(3)$ gives
$\left(\dfrac{3y_2 + 56}{7}\right)^2 + y_2^2 + 4 \times \left(\dfrac{3y_2 + 56}{7}\right) + 8y_2 - 38 = 0$
i.e. $\;$ $\dfrac{9y_2^2 + 336 y_2 + 3136}{49} + y_2^2 + \dfrac{12y_2 + 224}{7} + 8y_2 - 38 = 0$
i.e. $\;$ $9y_2^2 + 336 y_2 + 3136 + 49 y_2^2 + 84 y_2 + 1568 + 392 y_2 - 1862 = 0$
i.e. $\;$ $58 y_2^2 + 812 y_2 + 2842 = 0$
i.e. $\;$ $y_2^2 + 14y_2 + 49 = 0$
i.e. $\;$ $\left(y_2 + 7\right)^2 = 0$ $\implies$ $y_2 = -7$
Substituting the value of $y_2$ in equation $(4)$ gives
$7 x_2 - 3 \times \left(-7\right) - 56 = 0$
i.e. $\;$ $7 x_2 - 35 = 0$ $\implies$ $x_2 = 5$
$\therefore \;$ $Q \left(x_2, y_2\right) = \left(5, -7\right)$
Equation of tangent to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ at any given point $\left(x_1, y_1\right)$ is
$xx_1 + yy_1 + g \left(x + x_1\right) + f \left(y + y_1\right) + c = 0$
$\therefore \;$ Equation of tangent at $\left(-9, -1\right)$ is
$-9x - y + 2 \left(x - 9\right) + 4 \left(y - 1\right) - 38 = 0$
i.e. $\;$ $-9x - y + 2x -18 + 4y - 4 - 38 = 0$
i.e. $\;$ $7x - 3y + 60 = 0$
Equation of tangent at $\left(5, -7\right)$ is
$5x - 7y + 2 \left(x + 5\right) + 4 \left(y - 7\right) - 38 = 0$
i.e. $\;$ $5x - 7y + 2x + 10 + 4y - 28 - 38 = 0$
i.e. $\;$ $7x - 3y - 56 = 0$