
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
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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.
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.
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