Circles Part 1

$x^{2}+y^{2}+a x+2 a y+c=0(a<0)$ For a circle in standard form $x^{2}+y^{2}+2gx+2fy+c=0$ $x$-intercept $=2\sqrt{g^{2}-c}$ Here, $2g=a \Rightarrow g=\frac{a}{2}$ $2f=2a \Rightarrow f=a$ $x$-intercept $\Rightarrow 2\sqrt{\frac{a^{2}}{4}-c}=2\sqrt{2}$ $\Rightarrow \frac{a^{2}}{4}-c=2$.....(i) $y$-intercept $\Rightarrow 2\sqrt{a^{2}-c}=2\sqrt{5}$ $a^{2}-c=5$...(ii) Subtracting Eq. (ii) from Eq. (i), we get $\left(\frac{3}{4}\right)a^{2}=3$ $\Rightarrow a^{2}=4 \Rightarrow a=\pm 2$ As, $a<0$, so, $a=-2,\ c=-1$ So, $x^{2}+y^{2}-2x-4y-1=0$ $(x-1)^{2}+(y-2)^{2}-1=1+4$ $\Rightarrow (x-1)^{2}+(y-2)^{2}=6$ Centre $(1,2)$, Radius $=\sqrt{6}$ Equation of tangent is perpendicular to $x+2y=0$. So, equation of tangent will be $2x-y+\lambda=0$ Now, perpendicular distance of this line from $(1,2)$ will be $\sqrt{6}$ units. $\frac{\left|2\cdot1-2+\lambda\right|}{\sqrt{5}}=\sqrt{6}$ $\lambda=\pm\sqrt{30}$ So, $T\Rightarrow 2x-y\pm\sqrt{30}=0$ Now, shortest distance from origin $(0,0)$ to tangent $T$ will be $\frac{\left|2\cdot0-0\pm\sqrt{30}\right|}{\sqrt{2^{2}+1^{2}}}=\frac{\sqrt{30}}{\sqrt{5}}=\sqrt{6}$