Question

The length of the common chord of two circles ${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {c^2}{\text{ and }}{\left( {x - b} \right)^2} + {\left( {y - a} \right)^2} = {c^2}$ is
A. $\sqrt {4{c^2} + 2{{\left( {a - b} \right)}^2}} $
B. $\sqrt {4{c^2} + 2{{\left( {a + b} \right)}^2}} $
C. $\sqrt {4{c^2} - 2{{\left( {a - b} \right)}^2}} $
D. $\sqrt {{c^2} - 2{{\left( {a - b} \right)}^2}} $

Answer 1

Hint: If ${S_1}$ is the equation of first circle and ${S_2}$ is the equation of the second circle then the equation of common chord of both the circle is given as ${S_1} - {S_2} = 0$ and the distance from the point $\left( {{x_1},{y_1}} \right)$ on the line $ax - by = 0$ is given as $\left| {\dfrac{{a{x_1} - b{x_2}}}{{\sqrt {{a^2} + {b^2}} }}} \right|$.

Complete step by step answer:

The equation of first circle is
\[{S_1} = {\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {c^2}.............\left( 1 \right)\]
And equation of the second circle is
${S_2} = {\left( {x - b} \right)^2} + {\left( {y - a} \right)^2} = {c^2}...............\left( 2 \right)$
Then the center of circle ${S_1}$ is A (a, b) and radius is c.
The center of the circle of the circle ${S_2}$ is B (b, a) and radius is c.
Now, PQ is a common chord AB intersect PQ at M and
AP=BP=c
Equation of PQ is given by,
${S_1} - {S_2} = 0$
$
   \Rightarrow \left[ {{{\left( {x - a} \right)}^2} + {{\left( {y - b} \right)}^2} = {c^2}{\text{ }}} \right]{\text{ - }}\left[ {{{\left( {x - b} \right)}^2} + {{\left( {y - a} \right)}^2} = {c^2}} \right] \\
   \Rightarrow 2bx - 2by + 2ay - 2ax = 0 \\
   \Rightarrow 2b\left( {x - y} \right) + 2a\left( {y - x} \right) = 0 \\
   \Rightarrow 2b\left( {x - y} \right) - 2a\left( {x - y} \right) = 0 \\
   \Rightarrow \left( {x - y} \right)\left( {2b - 2a} \right) = 0 \\
   \Rightarrow x - y = 0 \\
$
Then equation of PQ is $x - y = 0$
AM = length of perpendicular from A (a, b) on PQ whose equation is $x - y = 0$
Therefore, AM $ = \dfrac{{\left| {1 \times a - 1 \times b} \right|}}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( { - 1} \right)}^2}} }} = \dfrac{{\left| {a - b} \right|}}{{\sqrt 2 }}$
In right $\Delta {\text{PMA,}}$
$
  PM = \sqrt {{{\left( {AP} \right)}^2} - {{\left( {AM} \right)}^2}} \\
  PM = \sqrt {{c^2} - \dfrac{{{{\left( {a - b} \right)}^2}}}{2}} \\
  \left[ {{\text{as AP = c }}\left( {{\text{radius of the circle}}} \right)} \right] \\
$
But M is the midpoint of PQ
Then, PQ =2PM = 2
$
  \sqrt {{c^2} - \dfrac{{{{\left( {a - b} \right)}^2}}}{2}} \\
   = \sqrt {4{c^2} - \dfrac{{4{{\left( {a - b} \right)}^2}}}{2}} \\
   = \sqrt {4{c^2} - 2{{\left( {a - b} \right)}^2}} \\
$

Hence, the correct option is “C”.

Note: In order to solve these types of problems remember all the formulas and equations of lines and circles. Draw the diagram of the given problem; this helps in solving the problem and visualization of the problem. Be familiar with the terms like length of the perpendicular, foot of the perpendicular, chord, tangent and many more.