Common Chord of two Circles: Equation, Properties, Formula

The common chord of two circles is that chord which joins each circle at two points. In a geometric sense, the endpoints of a chord lie on the circle. The common chord divides each circle into two segments. These two segments do not have to be equal in area unless the circles are of the same size and intersect symmetrically.

Common Chord of Two Circles

A circle is the locus of a moving point such that its distance from a fixed point is constant.

The fixed point is called the centre (O) of the circle and the constant distance is called its radius (r)

Equation of circle

The equation of a circle with centre at C (h,k) and radius r is (x - h)² + (y - k)² = r²

Using the distance formula, we have

$
\sqrt{(x-h)^2+(y-k)^2}=r
$

i.e.

$
(x-h)^2+(y-k)^2=r^2
$
If the centre of the circle is the origin or $(0,0)$ then the equation of the circle becomes

$
(x-0)^2+(y-0)^2=r^2
$

i.e. $x^2+y^2=r^2$

Equation of Common Chord of Two Circles

If two circles $S=0$ and $S^{\prime}=0$, intersect at two points let's say at $A$ and $B$. Then the equation of the line joining the points $A$ and $B$ is called the common chord of the two circles.

$
\begin{aligned}
& (\mathrm{S})=\mathrm{x}^2+\mathrm{y}^2+2 g \mathrm{~g}+2 \mathrm{fy}+\mathrm{c}=0 \\
& \left(S^{\prime}\right)=x^2+y^2+2 g^{\prime} x+2 f^{\prime} y+c^{\prime}=0 \text { is } \\
& 2 \mathrm{x}\left(\mathrm{g}-\mathbf{g}^{\prime}\right)+2 \mathrm{y}\left(\mathbf{f}-\mathbf{f}^{\prime}\right)+\mathbf{c}-\mathbf{c}^{\prime}=\mathbf{0} \\
& \text { or, } \mathbf{S}-\mathbf{S}^{\prime}=0
\end{aligned}
$

Length of Common Chord AB

$
\begin{aligned}
& \mathrm{AB}=2(\mathrm{AM}) \quad(\because \mathrm{M} \text { is the mid-point of } \mathrm{AB}) \\
& \quad=2 \sqrt{\left\{\left(\mathrm{C}_1 \mathrm{~A}\right)^2-\left(\mathrm{C}_1 \mathrm{M}\right)^2\right\}}
\end{aligned}
$

$\mathrm{C}_1 \mathrm{~A}=$ radius of the circle $\mathrm{S}=0$,
(i.e. $\mathrm{C}_1 \mathrm{~A}=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}$ )
$\mathrm{C}_1 \mathrm{M}=$ length of perpendicular from $\mathrm{C}_1$ on common chord AB
The length of common chord $A B$ of two circles is maximum when it is the diameter of the smaller circle between them.

Recommended Video Based on Equation of Common Chord of Two Circles

Solved Examples Based on Equation of Common Chord of Two Circles

Example 1: The common tangent to the circle $x^2+y^2=4$ and $x^2+y^2+6 x+8 y-24=0$ also passes through the point :
1) $(4,-2)$
2) $(-6,4)$
3) $(6,-2)$
4) $(-4,6)$
Solution:

The equation of the common chord of two circles
$(S)=x^2+y^2+2 g x+2 f y+c=0$
$\left(S^{\prime}\right)=x^2+y^2+2 g^{\prime} x+2 f^{\prime} y+c^{\prime}=0$ is
$2 \mathbf{x}\left(\mathrm{g}-\mathrm{g}^{\prime}\right)+2 \mathbf{y}\left(\mathrm{f}-\mathrm{f}^{\prime}\right)+\mathrm{c}-\mathrm{c}^{\prime}=0$
or, $\mathbf{S}-\mathbf{S}^{\prime}=\mathbf{0}$

$
\begin{aligned}
& x^2+y^2-4=0 \\
& x^2+y^2+6 x+8 y-24=0
\end{aligned}
$

common tangent will be $s_1-s_2=0$

$
\begin{aligned}
& 6 x+8 y=20 \\
& 3 x+4 y=10
\end{aligned}
$

Hence point $(6,-2)$ lies on the above line.

Example 2: If the circles $x^2+y^2+5 K x+2 y+K=0$ and $2\left(x^2+y^2\right)+2 K x+3 y-1=0(K \epsilon R)$, intersect at the points P and Q , then the line $4 x+5 y-K=0$ passes through P and Q , for:

1) infinitely many values of $K$
2) no value of $K$
3) exactly two values of $K$
4) exactly one value of $K$

Solution:

Common Chord of two Circles -

The equation of the common chord of two circles

$
\begin{aligned}
& (S)=x^2+y^2+2 g x+2 f y+c=0 \\
& \left(S^{\prime}\right)=x^2+y^2+2 g^{\prime} x+2 f^{\prime} y+c^{\prime}=0 \text { is } \\
& 2 \mathbf{x}\left(g-g^{\prime}\right)+2 \mathbf{y}\left(f-f^{\prime}\right)+\mathbf{c}-\mathbf{c}^{\prime}=\mathbf{0} \\
& \text { or, } \quad \mathbf{S}-\mathbf{S}^{\prime}=\mathbf{0}
\end{aligned}
$
Given two circles are

$
\begin{aligned}
& S_1=x^2+y^2+5 K x+2 y+K=0 \\
& S_2=2\left(x^2+y^2\right)+2 K x+3 y-1=0
\end{aligned}
$
Equation of common chord

$
\begin{aligned}
& S_1-S_2=0 \\
& \Rightarrow 4 k x+\frac{1}{2} y+k+\frac{1}{2}=0
\end{aligned}
$

$\qquad$
Given equation of chord is

$
4 x+5 y-k=0
$

$\qquad$
On Comparing (1) \& (2)

$
k=\frac{1}{10}=\frac{k+\frac{1}{2}}{-k}
$
There is no value of $k$
So, option (2) is correct.

Example 3: If the angle of intersection at a point where two circles with radii 5 cm and 12 cm intersects is $90^{\circ}$, then the length (in cm ) of their common chord is :

1)$\frac{13}{6}$

2) $\frac{13}{5}$
3) $\frac{120}{13}$
4) $\frac{60}{13}$

Solution:
Length of Common Chord -

$
\begin{aligned}
A B & =2(A M) \quad(\because m \text { is mid }- \text { point of } A B) \\
& =2 \sqrt{\left\{\left(C_1 A\right)^2-\left(C_1 M\right)^2\right\}}
\end{aligned}
$

$\mathrm{C}_1 \mathrm{~A}=$ radius of the circle $\mathrm{S}=0$, (i.e. $\mathrm{C}_1 \mathrm{~A}=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}$ )
$\mathrm{C}_1 \mathrm{M}=$ length of perpendicular from $\mathrm{C}_1$ on common chord $A B$

Length of common chord $=2 x$

$
\begin{aligned}
& \sqrt{s^2-x^2}+\sqrt{12^2-x^2}=13 \\
& x=\frac{12 \times 5}{13} \\
& 2 x=\frac{120}{13}
\end{aligned}
$
$
x^2+y^2+14 x
$
$
-4 y+28=0 \quad \text { and }
$

Example 4: If the lengths of external and internal common tangents to two circles $\quad x^2+y^2-14 x \quad+4 y-28=0$ are $\lambda$ and $\mu$ the value of $\lambda+\mu$ must be

1) 20

2) 25

3) 18

4) 24

Solution:

The given circles $S_1 \equiv x^2+y^2+14 x-4 y+28=0$ and $S_2 \equiv x^2+y^2-14 x+4 y-28=0$
Centres and radii of circles $S_1$ and $S_2$ are

$
C_1(-7,2), r_1=\sqrt{49+4-28}=5
$

and $C_2(7,-2), r_2=\sqrt{49+4+28}$
$=9$ respectively.

Here $d=C_1 C_2=\sqrt{(-7-7)^2+(2+2)^2}$

$
\begin{aligned}
& =\sqrt{212}>r_1+r_2=5+9=14 \\
\therefore \quad & d>r_1+r_2
\end{aligned}
$
Hence circles don't touch or cut
$\therefore$ Length of the external common tangent

$
\begin{aligned}
L_{e x} & =\sqrt{d^2-\left(r_2-r_1\right)^2} \\
& =\sqrt{212-(9-5)^2}=\sqrt{212-16} \\
& =\sqrt{196}=14=\lambda
\end{aligned}
$
and length of the internal common tangent

$
\begin{aligned}
& L_{i n}=\sqrt{d^2-\left(r_1+r_2\right)^2} \\
& =\sqrt{212-(5+9)^2}=\sqrt{212-196} \\
& =\sqrt{16}=4=\mu \\
& \therefore \lambda+\mu=18
\end{aligned}
$

Hence, the answer is option (3).

Example 5: Two circles in the first quadrant of radii $r_1$ and $r_2$ touch the coordinate axes. Each of them cuts off an intercept of 2 units with the line $x+y=2$. Then $r_1^2+r_2^2-r_1 r_2$ is equal to

1) 7

2) 8

3) 6

4) 5

Solution

$\begin{aligned} & \quad \operatorname{Circle}(x-a)^2+(y-a)^2=a^2 \\ & x^2+y^2-2 a x-2 a y+a^2=0 \\ & \text { Intercept }=2 \\ & \Rightarrow 2 \sqrt{a^2-d^2}=2\end{aligned}$

Where $\mathrm{d}=$ perpendicular distance of centre from line $x+y=2$

$
\begin{aligned}
& \Rightarrow 2 \sqrt{a^2-\left(\frac{a+a-2}{\sqrt{2}}\right)^2}=2 \\
& \Rightarrow a^2-\frac{(2 \mathrm{a}-2)^2}{2}=1 \Rightarrow 2 a^2-4 a^2+8 a-4=2 \\
& \Rightarrow 2 a^2-8 \mathrm{a}+6=0 \Rightarrow a^2-4 a+3=0 \\
& \therefore r_1+r_2=4 \text { and } r_1 r_2=3 \\
& \therefore r_1^2+r_2^2-r_1 r_2=\left(r_1+r_2\right)^2-3 r_1 r_2 \\
& =16-9=7
\end{aligned}
$