The common tangent to the circles x ^ 2 + y ^ 2 = 4 and x ^ 2 + y ^...

The centre and radius of the circle ${x}^{2} + {y}^{2} + 2gx + 2fy + c = 0$ are respectively $\left(- g, - f\right)$ and $\sqrt{{g}^{2} + {f}^{2} - c} .</p><p>Let, for the circle${S}_{1}: {x}^{2} + {y}^{2} = 4$centre${C}_{1}: \left(0, 0\right)$, {r}_{1} = 2$ units.

And, for the circle ${S}_{2}: {x}^{2} + {y}^{2} + 6x + 8y - 24 = 0$ centre

${C}_{2}\left(- 3, - 4\right)$, {r}_{2} = \sqrt{{3}^{2} + {4}^{2} + 24} = 7units.

The distance between the $points \left({x}_{1}, {y}_{1}\right) and \left({x}_{2}, {y}_{2}\right) is$\sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}}$</p><p>Hence, the distance between the centres of the circles is${C}_{1}{C}_{2} = \sqrt{{\left(- 3 - 0\right)}^{2} + {\left(- 4 - 0\right)}^{2}} = 5$units.</p><p>Also, \left|{r}_{1} - {r}_{2}\right| = 5$ units.

Since, ${C}_{1}{C}_{2} = \left|{r}_{1} - {r}_{2}\right|,$ hence both the circles touches each other internally.

Therefore, the common tangent is same as the radical axis of the two circles i.e. ${S}_{1} - {S}_{2} = 0$

Thus, the common tangent is ${x}^{2} + {y}^{2} - 4 - \left({x}^{2} + {y}^{2} + 6x + 8y - 24\right) = 0$

$⇒ - 6x - 8y + 20 = 0$

⇒ 3x + 4y - 10 = 0 .

Clearly, the $point \left(6, - 2\right) lies on the common tangent.</p>$