Question

Find the transverse common tangents of the circles:
$x^2+y^2-4x-10y+28=0$ and $x^2+y^2+4x-6y+4=0$

Answer 1

$x^2+y^2-4x-10y+28=0$

${(x-2)}^2+{(y-5)}^2=1^2$

$x^2+y^2+4x-6y+4=0$

${(x+2)}^2+{(y-3)}^2=3^2$

$\frac{AP}{BQ}=\frac{AO}{BO}=\frac{1}{3}$

So, $0=\frac{B+3A}{4}=\frac{(-2,3)+(6,15)}{4}$

$=(1,\frac{9}{2})$

$OB=\sqrt{3^2+{\left(\frac{3}{2}\right)}^2}=\frac{3}{2}\sqrt{5}$

So, $\sin \theta =\frac{QB}{OB}=\frac{3}{\frac{3}{2}\sqrt{5}}=\frac{2}{\sqrt{5}}$

$\tan \theta =\frac{\sin \theta }{\sqrt{1-{\sin }^2\theta }}=\frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}}=2$

Slope of $AB=\frac{3-5}{-2-2}=\frac{1}{2}$

So slope of tangents is $m=\frac{\frac{1}{2}\pm 2}{1\mp \frac{1}{2}\times 2}$

$=\infty or\frac{-3}{4}$

So, the tangents are $x=1$

or $y-\frac{9}{2}=\frac{-3}{4}(x-1)$

i.e. $4y-18=-3x+3$

$\Rightarrow 3x+4y=21$