Angle of Intersection of Two Circle: How to Find, Formula, Derivation

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 circle 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}
$
$
\begin{aligned}
& \text { 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) \\
& \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]}
\end{aligned}
$
$\cos \theta=\left|\frac{r_1^2+r_2^2-d^2}{2 r_1 r_2}\right|$

Note:
If the angle of the intersection of two circles is $90^{\circ}$, then the circles are said to be orthogonal circles.

The condition for orthogonality is $2\left(g_1 g_2+f_1 f_2\right)=c_1+c_2$

Recommended Video Based on Angle of Intersection of Two Circle

Solved Examples Based on Angle of Intersection of Two Circle

Example 1: The centre of circle S lies on $2 \mathrm{x}-2 \mathrm{y}+9=0$ and it cuts orthogonally the circle $x^2+y^2=4$. Then the circle passes through two fixed points
1) $(1,1),(3,3)$
2) $\left(-\frac{1}{2}, \frac{1}{2}\right),(-4,4)$
3) $(0,0),(5,5)$
4) none of these

Solution
Let $S=x^2+y^2+2 g x+2 f y+c=0$
$:$ it cuts $\mathrm{x}^2+\mathrm{y}^2=4_{\text {orthogonally }}$

$
\begin{aligned}
& \therefore \quad 2 g_1 g_2+2 f_1 f_2=c_1+c_2 \\
& \Rightarrow \quad c=4
\end{aligned}
$

$(-\mathrm{g},-\mathrm{f})$ lies on $2 \mathrm{x}-2 \mathrm{y}+9=0$

$
\begin{aligned}
& \therefore-2 g+2 f+9=0 \\
& \therefore \quad S \equiv x^2+y^2+2 g x+2 f y+4=0 \\
& \Rightarrow \quad x^2+y^2+(2 f+9) x+2 f y+4=0 \Rightarrow\left(x^2+y^2\right.
\end{aligned}
$
It is of the form $\mathrm{S}+\lambda \mathrm{P}=0$ and hence passes through the II $y$

Example 2: If a circle passes through the point $(\mathrm{a}, \mathrm{b})$ and cuts the circle $x^2+y^2=k^2$ orthogonally, equation of the locus of its centre is
1) $2 a x+2 b y=a^2+b^2+k^2$
2) $a x+b y=a^2+b^2+k^2$
3) $x^2+y^2+2 a x+2 b y+k^2=0$
4) $x^2+y^2-2 a x-2 b y+a^2+b^2-k^2=0$

Solution
Let the equation of the circle through $(\mathrm{a},(\mathrm{B})$ be

$
x^2+y^2+2 g x+2 f y+c=0
$

then $\mathrm{a}^2+\mathrm{b}^2+2 g \mathrm{a}+2 \mathrm{fb}+\mathrm{c}=0$
Since (i) cuts the circle $x^2+y^2=k^2$ orthogonally, we have

$
2 \mathrm{~g} \times 0+2 \mathrm{f} \times 0=\mathrm{c}-\mathrm{k}^2 \Rightarrow \mathrm{c}=\mathrm{k}^2
$

so that from (ii), we get $\mathrm{a}^2+\mathrm{b}^2+2 \mathrm{ga}+2 \mathrm{fb}+\mathrm{k}^2=0$, and the locus of the center of (i) is $2 a x+2 b y-\left(a^2+b^2+k^2\right)=0$

Hence, the answer is the option (1).

Example 3: The circles $x^2+y^2-10 x+9=0$ and $x^2+y^2=r^2$ intersect each other in two distinct points if
1) $r>8$
2) $\mathrm{r}<2$
3) $7<\mathrm{r}<11$
4) $1<r<9$

Solution

$
\begin{aligned}
& \mathrm{C}_1 \equiv(5,0), \mathrm{r}_1=4 \\
& \mathrm{C}_2 \equiv(0,0), \mathrm{r}_2=\mathrm{r} \\
& \left(\mathrm{C}_1 \mathrm{C}_2\right)=5
\end{aligned}
$
So, $r-4<5<r+4$

$
\Rightarrow \mathrm{r}<\mathrm{g} \& \mathrm{r}>1 \Rightarrow 1<\mathrm{r}<9
$
Hence, the answer is the option (4).

Example 4: If two circles $(\mathrm{x}-1)^2+(\mathrm{y}-3)^2=\mathrm{r}^2$ and $\mathrm{x}^2+\mathrm{y}^2-8 \mathrm{x}+2 \mathrm{y}+8=0$ intersect in two distinct points, then
1) $2<r<8$
2) $\mathrm{r}<2$
3) $r=2$
4) $r>2$

Solution
Centers and radii of the given circles are $\mathrm{C}_1(1,3), \mathrm{r}_1=\mathrm{r}$ and $\overline{\mathrm{C}_2}=(4,-1), \mathrm{r}_2=3$ respectively since circles intersect in two distinct points, then

$
\begin{aligned}
& \left|\mathrm{r}_1-\mathrm{r}_2\right|<\mathrm{C}_1 \mathrm{C}_2<\mathrm{r}_1+\mathrm{r}_2 \\
& \Rightarrow|\mathrm{r}-3|<5<\mathrm{r}+3
\end{aligned}
$

from the last two relations, $r>2$
from first two relations

$
\begin{aligned}
& |\mathrm{r}-3|<5 \Rightarrow-5<\mathrm{r}-3<5 \\
& \Rightarrow-2<\mathrm{r}<8
\end{aligned}
$

from eqs. (i) and (ii), we get $2<\mathrm{r}<8$
Hence, the answer is the option (1).

Example 5: The locus of the centres of the circles which cut the circles $x^2+y^2+4 x-6 y+9=0$ and $x^2+y^2-5 x+4 y-2=0$ orthogonally is
1) $9 x+10 y-7=0$
2) $x-y+2=0$
3) $9 x-10 y+11=0$
4) $9 x+10 y+7=0$

Solution

[Hint: Locus of the centre of the $\odot$ cutting $\mathrm{S}_1=0$ and $\mathrm{S}_2=0$ orthogonally is the radical axis between $\mathrm{S}_1=0$ and $\mathrm{S}_2=0$ ]

| Let out circle be $x^2+y^2+2 g x+2 f y+c=0$ conditions $2(-\mathrm{g})(-2)+2(-\mathrm{f})(3)=\mathrm{c}+9$ and $2(-\mathrm{g})(5 / 2)+2(-\mathrm{f})(-2)=\mathrm{c}-2$ $\therefore \mathrm{ag}-10 \mathrm{f}=11$
$\therefore$ locus of centre $9 \mathrm{x}-10 \mathrm{y}+11=0$ Hence, the answer is the option(3).

Summary

The angle of intersection between two circles provides valuable insights into the geometric relationship between them. By using the formulas and understanding the properties of the circles, one can determine the angle formed by the tangents at the points of intersection. This concept is applicable in various fields, including engineering, computer graphics, astronomy, and mathematical problem-solving.