Find the equation of the circle which has extremities of a diameter the origin and the point $\left(2, -4\right)$.
Find also the equations of the tangents to the circle which are parallel to this diameter.
The extremities of the diameter are
origin, $\;$ i.e. $\;$ $\left(x_1, y_1\right) = \left(0, 0\right)$ $\;$ and $\;$ $\left(x_2, y_2\right) = \left(2, -4\right)$
$\therefore \;$ Equation of required circle is
$\left(x - x_1\right) \left(x - x_2\right) + \left(y - y_1\right) \left(y - y_2\right) = 0$
i.e. $\;$ $\left(x - 0\right) \left(x - 2\right) + \left(y - 0\right) \left(y + 4\right) = 0$
i.e. $\;$ $x^2 + y^2 - 2x + 4y = 0$ $\;\;\; \cdots \; (1)$
Slope of the line passing through the points $\;$ $\left(0, 0\right)$ $\;$ and $\;$ $\left(2, -4\right)$ $\;$ is
$m = \dfrac{-4}{2} = -2$
Since the required tangents are parallel to the line through $\left(0, 0\right)$ and $\left(2, -4\right)$,
$\therefore \;$ Slope of required tangents $= m = -2$
Comparing equation $(1)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives
$g = -1, \; f = 2, \; c = 0$
$\therefore \;$ Radius of circle $(1)$ is
$a = \sqrt{g^2 + f^2 - c} = \sqrt{\left(-1\right)^2 + 2^2 - 0} = \sqrt{5}$
Equations of required tangents are
$y = mx \pm a \sqrt{1 + m^2}$
i.e. $\;$ $y = -2x \pm \sqrt{5} \times \sqrt{1 + \left(-2\right)^2}$
i.e. $\;$ $y = -2x \pm 5$
$\therefore \;$ The equations of the required tangents are
$2x + y + 5 = 0$ $\;$ and $\;$ $2x + y - 5 = 0$