Angle of Intersection of Two Circle - Practice Questions & MCQ

The angle of the Intersection of Two Circle

The angle of Intersection of Two circles is defined as the angle between the tangents drawn to both circles at their point of intersection.

Let the equation of two circles be

$
\begin{aligned}
& S_1: x^2+y^2+2 g_1 x+2 f_1 y+c_1=0 \\
& S_2: x^2+y^2+2 g_2 x+2 f_2 y+c_2=0
\end{aligned}
$

$\mathrm{C}_1$ and $\mathrm{C}_2$ are the centres of the given circles and $r_1$ and $r_2$ are the radii of the circles.

Thus $C_1=\left(-g_1,-f_1\right)$ and $C_2=\left(-g_2,-f_2\right)$

$
r_1=\sqrt{g_1^2+f_1^2-c_1}
$

and $\quad r_2=\sqrt{g_2^2+f_2^2-c_2}$
Let

$
\begin{aligned}
d=C_1 C_2 & =\sqrt{\left(g_1-g_2\right)^2+\left(f_1-f_2\right)^2} \\
& =\sqrt{g_1^2+g_2^2+f_1^2+f_2^2-2\left(g_1 g_2+f_1 f_2\right)}
\end{aligned}
$
In $\Delta C_1 P C_2, \quad \cos \alpha=\left(\frac{r_1^2+r_2^2-d^2}{2 r_1 r_2}\right)$

$
\begin{gathered}
\Rightarrow \quad \cos \left(180^{\circ}-\theta\right)=\left(\frac{r_1^2+r_2^2-d^2}{2 r_1 r_2}\right) \\
{\left[\because \quad \alpha+\theta+90^{\circ}+90^{\circ}=360^{\circ}\right]} \\
\quad \cos \theta=\left|\frac{\mathrm{r}_1^2+\mathrm{r}_2^2-\mathrm{d}^2}{2 \mathrm{r}_1 \mathrm{r}_2}\right|
\end{gathered}
$

For finding $\cos \theta$, we are using COSINE FORMULA