Question

The length of the common chord of two circles ${(x-a)}^2+{(y-b)}^2=c^{2}and{(x-b)}^2+{(y-a)}^2=c^2$ is

• A. $\sqrt{{4c}^2+2{(a-b)}^2}$
• B. $\sqrt{{4c}^2+2{(a+b)}^2}$
• C. $\sqrt{{4c}^2-2{(a-b)}^2}$
• D. $\sqrt{c^2-2{(a-b)}^2}$

Answer 1

The correct option is C $\sqrt{{4c}^2-2{(a-b)}^2}$

R.E.F image

$S_1=(x-a{)}^2+(y-b{)}^2=c^2$

$S_2=(x-b{)}^2+(y-a{)}^2=c^2$

$\therefore E{q}^n$ of common chord

$\to S_1-S_2=0$

$\Rightarrow (x-a{)}^2+(y-b{)}^2-(x-b{)}^2-(y-a{)}^2=0$

$\Rightarrow -2xa-2yb+2xb+2ya=0$

$\Rightarrow (a-b)(y-x)=0$

$e{q}^n$ of common chord $\Rightarrow y=x$

$C_{1}M=$ length of $\perp$ from $C_1(a,b)$ an line $PQx-y=0$

Length $C_{1}M=\frac{|a-b|}{2}$

$C_{1}P=$ radius of $1^{st}$ circle = C

$\therefore$ In $\Delta P{C}_{1}M,PM=\sqrt{(P{C}_1{)}^2+(C_{1}M{)}^2}$

$\Rightarrow \sqrt{c^2-\frac{(a-b{)}^2}{2}}$

$PQ=2PM=2\sqrt{C^2-\frac{(a-b{)}^2}{2}}$

$PQ=\sqrt{4C^2-2(a-b{)}^2}$ (option C)