What is the equation of the perpendicular bisector of a chord of a circle?
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Hint: Here in this question we have been asked for the equation of the perpendicular bisector of a chord of a circle for answering this question we will assume two end points of a chord of a circle and find its midpoint through which the perpendicular bisector will pass.
Complete step-by-step answer:
Now considering from the question we have been asked for the equation of the perpendicular bisector of a chord of a circle.
Let us assume a chord $AB$ with $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ then find its perpendicular bisector.
From the basic concepts we know that the midpoint of the line formed by the points $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ will be given as $\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$ .
The midpoint of the chord $AB$ will be given as $M\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$ .
From the basic concepts we know that the slope of the line formed by the points $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ will be given as $\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ .
The slope of the chord $AB$ will be given as $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ .
From the basic concepts we know that the product of slopes of two perpendicular lines will be given as $-1$ .
Now we can say that the perpendicular of the chord will pass through the midpoint and has the slope $\dfrac{-1}{m}$ .
From the basic concepts we know that the line of an equation passing through a point $\left( p,q \right)$ and having slope $k$ is generally given as $y-q=k\left( x-p \right)$ .
Hence we can say that the equation of perpendicular bisector will be given as $y-\dfrac{{{y}_{1}}+{{y}_{2}}}{2}=\left( \dfrac{{{x}_{1}}-{{x}_{2}}}{{{y}_{2}}-{{y}_{1}}} \right)\left( x-\dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right)$ .
Therefore we can conclude that the equation of the perpendicular bisector of a chord of a circle will be given as $y-\dfrac{{{y}_{1}}+{{y}_{2}}}{2}=\left( \dfrac{{{x}_{1}}-{{x}_{2}}}{{{y}_{2}}-{{y}_{1}}} \right)\left( x-\dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right)$ .
Note: In the process of answering questions of this type we should be sure with the concepts that we are going to apply in between the steps. Someone can make mistakes if they have some confusion in the concept.
Complete step-by-step answer:
Now considering from the question we have been asked for the equation of the perpendicular bisector of a chord of a circle.
Let us assume a chord $AB$ with $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ then find its perpendicular bisector.
From the basic concepts we know that the midpoint of the line formed by the points $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ will be given as $\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$ .
The midpoint of the chord $AB$ will be given as $M\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$ .
From the basic concepts we know that the slope of the line formed by the points $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ will be given as $\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ .
The slope of the chord $AB$ will be given as $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ .
From the basic concepts we know that the product of slopes of two perpendicular lines will be given as $-1$ .
Now we can say that the perpendicular of the chord will pass through the midpoint and has the slope $\dfrac{-1}{m}$ .
From the basic concepts we know that the line of an equation passing through a point $\left( p,q \right)$ and having slope $k$ is generally given as $y-q=k\left( x-p \right)$ .
Hence we can say that the equation of perpendicular bisector will be given as $y-\dfrac{{{y}_{1}}+{{y}_{2}}}{2}=\left( \dfrac{{{x}_{1}}-{{x}_{2}}}{{{y}_{2}}-{{y}_{1}}} \right)\left( x-\dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right)$ .
Therefore we can conclude that the equation of the perpendicular bisector of a chord of a circle will be given as $y-\dfrac{{{y}_{1}}+{{y}_{2}}}{2}=\left( \dfrac{{{x}_{1}}-{{x}_{2}}}{{{y}_{2}}-{{y}_{1}}} \right)\left( x-\dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right)$ .
Note: In the process of answering questions of this type we should be sure with the concepts that we are going to apply in between the steps. Someone can make mistakes if they have some confusion in the concept.
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