Q.Show that the circles $x^2+y^2-36=0$ and $x^2+y^2-6x-8y+24=0$ touch each other internally, and find the point of contact and the common tangent at that point.

The angle between two intersecting circles is found from $\cos\theta = \dfrac{r_1^2+r_2^2-d^2}{2r_1r_2}$, where $d$ is the distance between centres.

Circle 1: $x^2+y^2+6x-10y-135=0 \Rightarrow g_1=3, f_1=-5, c_1=-135$

$r_1=\sqrt{9+25+135}=\sqrt{169}=13,\quad C_1=(-3,5)$

Circle 2: $x^2+y^2-4x+14y-116=0 \Rightarrow g_2=-2, f_2=7, c_2=-116$

$r_2=\sqrt{4+49+116}=\sqrt{169}=13,\quad C_2=(2,-7)$

Distance between centres:

$d^2=(-3-2)^2+(5-(-7))^2=25+144=169$

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Use the angle-between-two-circles formula $\cos\theta=\dfrac{r_1^2+r_2^2-d^2}{2r_1r_2}$ with the given circles' centres/radii.

Circle 1: $x^2+y^2-12x-6y+41=0$ has centre $(6,3)$ and $r_1^2=36+9-41=4$, so $r_1=2$.

Circle 2: $x^2+y^2+kx+6y-59=0$ has centre $(-k/2,-3)$ and $r_2^2=\dfrac{k^2}{4}+9+59=\dfrac{k^2}{4}+68$.

Distance squared between centres: $d^2=\left(6+\dfrac{k}{2}\right)^2+36$.

$r_1^2+r_2^2-d^2 = 4+\dfrac{k^2}{4}+68-\left(6+\dfrac{k}{2}\right)^2-36 = -6k$ (the $k^2$ terms and constants cancel).

Using $\cos45^\circ=\dfrac{1}{\sqrt2}$:

$\dfrac{-6k}{2(2)r_2}=\dfrac{1}{\sqrt2}\Rightarrow -6k\sqrt2=4r_2$.

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