Question

If a$\ne$b then the length of common chord of the circles ${\left(x-a\right)}^2+{\left(y-b\right)}^2=c^2$ and ${\left(x-b^{}\right)}^2+{\left(y-a\right)}^2=c^2$ is

• A. $\sqrt{c^2-{\left(a-b\right)}^2}$
• B. $\sqrt{4c^2-2{\left(a-b\right)}^2}$
  
• C. $\sqrt{3c^2-{\left(a-b\right)}^2}$
• D. $\sqrt{2c^2-{\left(a-b\right)}^2}$

Answer 1

Answer: (B)

$\sqrt{4c^2-2{\left(a-b\right)}^2}$

We have,

${(x-a)}^2+{(y-b)}^2=c^2$
$S_1\equiv x^2+a^2-2ax+y^2+b^2-2by-c^2=0$ …….. (1)

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

$S_2\equiv x^2+b^2-2bx+y^2+a^2-2ay-c^2=0$ ……… (2)

Since, $a\ne b$

Centre of the circle $S_1=(a,b)$ and radius $r_1=c$.

We know that the equation of common chord is $S_1-S_2=0$

So, the equation is

$(b-a)x+(a-b)y=0$ …….. (3)

We know that the length of common chord is

$=2\sqrt{{r_1}^2-{d_1}^2}$ ………. (4)

Where $r_1=$ radius and $d_1$ is the length of perpendicular drawn from the centre to the chord.

So,

$d_1=\left|\frac{(b-a)a+(a-b)b}{\sqrt{{(b-a)}^2+{(a-b)}^2}}\right|$

$d_1=\left|\frac{ab-a^2+ab-b^2}{\sqrt{{(a-b)}^2+{(a-b)}^2}}\right|$

$d_1=\left|\frac{2ab-a^2-b^2}{\sqrt{2{(a-b)}^2}}\right|$

$d_1=\left|\frac{-{(a-b)}^2}{\sqrt{2{(a-b)}^2}}\right|$

$d_1=\left|\frac{-(a-b)}{\sqrt{2}}\right|$

$d_1=\frac{(a-b)}{\sqrt{2}}$

From equation (4),

The length of common chord $=2\sqrt{c^2-{\left(\frac{a-b}{\sqrt{2}}\right)}^2}$

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

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

$=\sqrt{{\frac{8c^2-4(a-b)}{2}}^2}$

$=\sqrt{4c^2-2{(a-b)}^2}$

Hence, this is the answer.