Let the lengths of intercepts on x-axis and y-axis made by the circle
x² + y² + ax + 2ay + c = 0, (a < 0) be 2${\sqrt 2 }$ and 2${\sqrt 5 }$, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line x + 2y = 0, is equal to :

$2\sqrt {{{{a^2}} \over 4} - c} = 2\sqrt 2$<br><br>$\sqrt {{a^2} - 4c} = 2\sqrt 2$<br><br>${a^2} - 4c = 8$ .... (1)<br><br>$2\sqrt {{a^2} - c} = 2\sqrt 5$<br><br>${a^2} - c = 5$ .... (2)<br><br>$(2) - (1)$<br><br>$3c = - 3a \Rightarrow c = - 1$<br><br>${a^2} = 4 \Rightarrow a = - 2$ (Given a &lt; 0)<br><br>Equation of circle <br><br>${x^2} + {y^2} - 2x - 4y - 1 = 0$<br><br>Equation of tangent which is perpendicular to the line x + 2y = 0 is <br><br>$2x - y + \lambda = 0$<br><br>$\therefore$ p = r<br><br>$\left| {{{2 - 2 + \lambda } \over {\sqrt 5 }}} \right| = \sqrt 6$<br><br>$\Rightarrow \lambda = \pm \sqrt {30}$<br><br>$\therefore$ Tangent $2x - y \pm \sqrt {30} = 0$<br><br>Distance from origin = ${{\sqrt {30} } \over {\sqrt 5 }} = \sqrt 6$