Question

Find the direct common tangents of the circles $x^2+y^2+22x-4y-100=0$ and $x^2+y^2-22x+4y+100=0$.

Answer 1

For the circle

$x^2+y^2+22x-4y-100=0$

Center$=C_1(-11,2)$

radius$=\sqrt{(-11{)}^2+2^2+100}=\sqrt{121+4+100}=15$

For the circle

$x^2+y^2-22x+4y+100=0$

Center$=C_2(11,-2)$

radius$=\sqrt{(11{)}^2+(-2{)}^2+100}=\sqrt{121+4-100}=5$

$C_1{C}_2=\sqrt{(11+11{)}^2+(-2-2{)}^2}=10\sqrt{5}>(r_1+r_2)=10\sqrt{5}>\left(20\right)$

Now, taking

$\sin \theta =\frac{1}{\sqrt{5}}=\frac{5}{O{C}_2}\Rightarrow O{C}_2=5\sqrt{5}$

now, by using section formula with the ratio of $2:1$

$(\frac{2h-11}{3},\frac{2k+2}{3})=(11,-2)$

Solving we get, $h=22,k=-4$

So, slope$=\frac{y_2-y_1}{x_2-x_1}=\frac{4}{-22}=m$

now,

$m=\tan \theta =|\frac{m+\frac{2}{11}}{1-\frac{2m}{11}}|=\frac{1}{2}$

taking the case of,

$m=\tan \theta =\frac{m+\frac{2}{11}}{1-\frac{2m}{11}}=\frac{1}{2}$

$m=\frac{7}{24}$

taking the case of,

$m=\tan \theta =\frac{m+\frac{2}{11}}{1-\frac{2m}{11}}=\frac{-1}{2}$

$m=\frac{-3}{4}$

So, the equations would be

$y+4=m(x-22)$

taking $m=\frac{7}{24}$

$y+4=\frac{7}{24}(x-22)$
$7x-24y-250=0$

taking $m=\frac{-3}{4}$

$y+4=\frac{-3}{4}(x-22)$

$3x+4y-50=0$