Question

Find the tangent of the acute angle between the tangents at $(2,-5)$ to the circles $x^2+y^2=29$ and $x^2+y^2-8x-13=0$.

Answer 1

$S_1\equiv x^2+y^2=29;S_2\equiv x^2+y^2-8x-13=0$

For $S_1$: Slope of tangent:

$2x+2y\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=\frac{-x}{y}\Rightarrow {\left(\frac{dy}{dx}\right)}_{(2,-5)},\frac{2}{5}=m_1$

For $S_2$:

$2x+2y\frac{dy}{dx}-8=0$

$\Rightarrow \frac{dy}{dx}=\frac{8-2x}{2y}=\frac{4-x}{y}$

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(2,-5)}=\frac{-2}{5}$

$\tan \theta =\frac{m_1-m_2}{1+m_1{m}_2}=\frac{\frac{4}{5}}{1-\frac{4}{25}}=\frac{\frac{4}{5}}{\frac{21}{25}}=\frac{4}{5}\times \frac{25}{21}$

$=\frac{20}{21}$