The common tangent to the circles x 2 + y 2 = 4 and x 2 + y 2 + 6 x + 8 y - 24 = 0 also passes through the…

The centre and radius of the circle $x^2+y^2+2gx+2fy+c=0$ are respectively $\left(-g, -f\right)$ and $\sqrt{g^2+f^2-c}.$

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

$C_2\left(-3, -4\right), r_2=\sqrt{3^2+4^2+24}=7$ units.

The distance between the points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$

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.

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 $x^2+y^2-4-\left(x^2+y^2+6x+8y-24\right)=0$

$\Rightarrow -6x-8y+20=0$

$\Rightarrow 3x+4y-10=0.$

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