Prove that equation $r=a\cos \left( \theta -\alpha \right)\text{ and }r=a\sin \left( \theta -\alpha \right)$ represent two circles which cut at right angles.
Answer
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Hint: To prove equation $r=a\cos \left( \theta -\alpha \right)\text{ and }r=a\sin \left( \theta -\alpha \right)$ represent circles, we will convert them to Cartesian coordinate and then obtain equation of circle from which we can find center and radius of both circles. After that, we will use condition of two circles to cut at right angles to proof that they cut at right angles. Condition is given by:
\[{{\left( \text{Distance between centers} \right)}^{2}}=\text{Radius}_{1}^{2}+\text{Radius}_{2}^{2}\]
Where, $\text{Radiu}{{\text{s}}_{1}}$ is radius of first circle and $\text{Radiu}{{\text{s}}_{2}}$ is radius of second circle.
Complete step by step answer:
Let us first consider first equation which is $r=a\cos \left( \theta -\alpha \right)$
Using $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$ in above equation to simplify, we get:
\[r=a\cos \theta \cos \alpha +a\sin \theta \sin \alpha \cdots \cdots \cdots \cdots \left( i \right)\]
As we can see, equation is in polar coordinates, we will convert them into Cartesian coordinate to find center and radius of circle it represents. Let us substitute:
\[x=r\cos \theta \text{ and }y=r\sin \theta \cdots \cdots \cdots \cdots \left( ii \right)\]
Finding $\cos \theta \text{ and }\sin \theta $ from above equations, we get:
\[\cos \theta =\dfrac{x}{r}\text{ and }\sin \theta =\dfrac{y}{r}\cdots \cdots \cdots \cdots \left( iii \right)\]
Squaring and adding (ii) we get:
\[{{x}^{2}}+{{y}^{2}}={{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)\]
As we know, ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ hence
\[{{x}^{2}}+{{y}^{2}}={{r}^{2}}\cdots \cdots \cdots \cdots \left( iv \right)\]
Now, putting (iii) in (i) we get:
\[\begin{align}
& r=a\left( \dfrac{x}{r} \right)\cos \alpha +a\left( \dfrac{y}{r} \right)\sin \alpha \\
& \Rightarrow {{r}^{2}}=ax\cos \alpha +ay\sin \alpha \\
\end{align}\]
Putting (iv) in above equation we get:
\[\begin{align}
& {{x}^{2}}+{{y}^{2}}=ax\cos \alpha +ay\sin \alpha \\
& \Rightarrow {{x}^{2}}-ax\cos \alpha +{{y}^{2}}-ay\sin \alpha =0 \\
\end{align}\]
To convert it into equation of circle, we will complete squares of x and y by adding and subtracting ${{\left( \dfrac{a}{2} \right)}^{2}}{{\cos }^{2}}\alpha \text{ and }{{\left( \dfrac{a}{2} \right)}^{2}}{{\sin }^{2}}\alpha $ in above equation, we get:
\[\begin{align}
& {{x}^{2}}-ax\cos \alpha +{{\left( \dfrac{a}{2} \right)}^{2}}{{\cos }^{2}}\alpha -{{\left( \dfrac{a}{2} \right)}^{2}}{{\cos }^{2}}\alpha +{{y}^{2}}-ay\sin \alpha +{{\left( \dfrac{a}{2} \right)}^{2}}{{\sin }^{2}}\alpha -{{\left( \dfrac{a}{2} \right)}^{2}}{{\sin }^{2}}\alpha =0 \\
& \Rightarrow {{\left( x-\dfrac{a}{2}\cos \alpha \right)}^{2}}+{{\left( y-\dfrac{a}{2}\sin \alpha \right)}^{2}}={{\left( \dfrac{a}{2} \right)}^{2}}\left( {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha \right) \\
\end{align}\]
We know ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$ therefore,
\[{{\left( x-\dfrac{a}{2}\cos \alpha \right)}^{2}}+{{\left( y-\dfrac{a}{2}\sin \alpha \right)}^{2}}={{\left( \dfrac{a}{2} \right)}^{2}}\]
As we know general equation of circle is given by ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ where (h, k) gives center of circle and r gives radius of circle. Comparing above equation with ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ we get that
\[\text{Center}=\left( \dfrac{a}{2}\cos \alpha ,\dfrac{a}{2}\sin \alpha \right)\text{ and Radius}=\dfrac{a}{2}\]
Using same procedure on second equation, consider second equation $r=a\sin \left( \theta -\alpha \right)$
Using $\sin \left( \theta -\alpha \right)=\sin \theta \cos \alpha -\cos \theta \sin \alpha $ we get $r=a\sin \theta \cos \alpha +a\cos \theta \sin \alpha $ converting to Cartesian coordinate by substituting $x=r\cos \theta \text{ and }y=r\sin \theta $ we get:
\[\cos \theta =\dfrac{x}{r}\text{ and }\sin \theta =\dfrac{y}{r}\]
Putting them in above equation, we get:
\[\begin{align}
& r=\dfrac{a}{r}y\cos \alpha -\dfrac{a}{r}x\sin \alpha \\
& \Rightarrow {{r}^{2}}=ay\cos \alpha -ax\sin \alpha \\
\end{align}\]
Using ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$ in above equation, we get:
\[\begin{align}
& {{x}^{2}}+{{y}^{2}}=ay\cos \alpha -ax\sin \alpha \\
& \Rightarrow {{x}^{2}}-ay\cos \alpha +{{y}^{2}}+ax\sin \alpha =0 \\
\end{align}\]
Completing squares of x and y by adding and subtracting ${{\left( \dfrac{a}{2} \right)}^{2}}{{\cos }^{2}}\alpha \text{ and }{{\left( \dfrac{a}{2} \right)}^{2}}{{\sin }^{2}}\alpha $ we get:
\[\begin{align}
& {{x}^{2}}-ay\cos \alpha +{{\left( \dfrac{a}{2} \right)}^{2}}{{\cos }^{2}}\alpha -{{\left( \dfrac{a}{2} \right)}^{2}}{{\cos }^{2}}\alpha +{{y}^{2}}-ax\sin \alpha +{{\left( \dfrac{a}{2} \right)}^{2}}{{\sin }^{2}}\alpha -{{\left( \dfrac{a}{2} \right)}^{2}}{{\sin }^{2}}\alpha =0 \\
& \Rightarrow {{\left( x+\dfrac{a}{2}\sin \alpha \right)}^{2}}+{{\left( y-\dfrac{a}{2}\cos \alpha \right)}^{2}}={{\left( \dfrac{a}{2} \right)}^{2}} \\
\end{align}\]
Comparing above equation by general equation of circle ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ we find that center of above circle is $\left( -\dfrac{a}{2}\sin \alpha ,\dfrac{a}{2}\cos \alpha \right)$ and radius is $\dfrac{a}{2}$.
Hence, we have to find center and radius of both circles. Now we will apply condition to check if both circles cut at right angle which is
\[{{\left( \text{Distance between centers} \right)}^{2}}=\text{Radius}_{1}^{2}+\text{Radius}_{2}^{2}\]
Distance between two points $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ is given by:
\[\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}\]
Applying formula, distance between centers $\left( \dfrac{a}{2}\cos \alpha ,\dfrac{a}{2}\sin \alpha \right)\text{ and }\left( -\dfrac{a}{2}\sin \alpha ,\dfrac{a}{2}\cos \alpha \right)$ is given by
\[\begin{align}
& \sqrt{{{\left( \dfrac{a}{2}\cos \alpha ,\dfrac{a}{2}\sin \alpha \right)}^{2}}+{{\left( -\dfrac{a}{2}\sin \alpha ,\dfrac{a}{2}\cos \alpha \right)}^{2}}} \\
& \Rightarrow \sqrt{\dfrac{{{a}^{2}}}{4}{{\cos }^{2}}\alpha +\dfrac{{{a}^{2}}}{4}{{\sin }^{2}}\alpha +\dfrac{{{a}^{2}}}{2}\sin \alpha \cos \alpha +\dfrac{{{a}^{2}}}{4}{{\sin }^{2}}\alpha +\dfrac{{{a}^{2}}}{4}{{\cos }^{2}}\alpha -\dfrac{{{a}^{2}}}{2}\sin \alpha \cos \alpha } \\
\end{align}\]
Using ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$ we get
\[\begin{align}
& \sqrt{\dfrac{{{a}^{2}}}{4}+\dfrac{{{a}^{2}}}{2}\sin \alpha \cos \alpha +\dfrac{{{a}^{2}}}{4}-\dfrac{{{a}^{2}}}{2}\sin \alpha \cos \alpha } \\
& \Rightarrow \sqrt{\dfrac{{{a}^{2}}}{2}} \\
\end{align}\]
Radius of first circle \[\Rightarrow \text{Radiu}{{\text{s}}_{\text{1}}}=\dfrac{a}{2}\]
Radius of second circle \[\Rightarrow \text{Radiu}{{\text{s}}_{\text{2}}}=\dfrac{a}{2}\]
Putting all of them in ${{\left( \text{Distance between centers} \right)}^{2}}=\text{Radius}_{1}^{2}+\text{Radius}_{2}^{2}$ we get:
\[\begin{align}
& {{\left( \sqrt{\dfrac{{{a}^{2}}}{2}} \right)}^{2}}={{\left( \dfrac{a}{2} \right)}^{2}}+{{\left( \dfrac{a}{2} \right)}^{2}} \\
& \Rightarrow \dfrac{{{a}^{2}}}{2}=\dfrac{{{a}^{2}}}{4}+\dfrac{{{a}^{2}}}{2} \\
\end{align}\]
Therefore, $\dfrac{{{a}^{2}}}{2}=\dfrac{{{a}^{2}}}{2}$
Hence, two circles cut each other at right angle.
Note:
Students should take care while converting polar coordinates to Cartesian coordinates. Students should also learn these formulas and conditions and apply them carefully while solving these types of sums.
\[{{\left( \text{Distance between centers} \right)}^{2}}=\text{Radius}_{1}^{2}+\text{Radius}_{2}^{2}\]
Where, $\text{Radiu}{{\text{s}}_{1}}$ is radius of first circle and $\text{Radiu}{{\text{s}}_{2}}$ is radius of second circle.
Complete step by step answer:
Let us first consider first equation which is $r=a\cos \left( \theta -\alpha \right)$
Using $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$ in above equation to simplify, we get:
\[r=a\cos \theta \cos \alpha +a\sin \theta \sin \alpha \cdots \cdots \cdots \cdots \left( i \right)\]
As we can see, equation is in polar coordinates, we will convert them into Cartesian coordinate to find center and radius of circle it represents. Let us substitute:
\[x=r\cos \theta \text{ and }y=r\sin \theta \cdots \cdots \cdots \cdots \left( ii \right)\]
Finding $\cos \theta \text{ and }\sin \theta $ from above equations, we get:
\[\cos \theta =\dfrac{x}{r}\text{ and }\sin \theta =\dfrac{y}{r}\cdots \cdots \cdots \cdots \left( iii \right)\]
Squaring and adding (ii) we get:
\[{{x}^{2}}+{{y}^{2}}={{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)\]
As we know, ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ hence
\[{{x}^{2}}+{{y}^{2}}={{r}^{2}}\cdots \cdots \cdots \cdots \left( iv \right)\]
Now, putting (iii) in (i) we get:
\[\begin{align}
& r=a\left( \dfrac{x}{r} \right)\cos \alpha +a\left( \dfrac{y}{r} \right)\sin \alpha \\
& \Rightarrow {{r}^{2}}=ax\cos \alpha +ay\sin \alpha \\
\end{align}\]
Putting (iv) in above equation we get:
\[\begin{align}
& {{x}^{2}}+{{y}^{2}}=ax\cos \alpha +ay\sin \alpha \\
& \Rightarrow {{x}^{2}}-ax\cos \alpha +{{y}^{2}}-ay\sin \alpha =0 \\
\end{align}\]
To convert it into equation of circle, we will complete squares of x and y by adding and subtracting ${{\left( \dfrac{a}{2} \right)}^{2}}{{\cos }^{2}}\alpha \text{ and }{{\left( \dfrac{a}{2} \right)}^{2}}{{\sin }^{2}}\alpha $ in above equation, we get:
\[\begin{align}
& {{x}^{2}}-ax\cos \alpha +{{\left( \dfrac{a}{2} \right)}^{2}}{{\cos }^{2}}\alpha -{{\left( \dfrac{a}{2} \right)}^{2}}{{\cos }^{2}}\alpha +{{y}^{2}}-ay\sin \alpha +{{\left( \dfrac{a}{2} \right)}^{2}}{{\sin }^{2}}\alpha -{{\left( \dfrac{a}{2} \right)}^{2}}{{\sin }^{2}}\alpha =0 \\
& \Rightarrow {{\left( x-\dfrac{a}{2}\cos \alpha \right)}^{2}}+{{\left( y-\dfrac{a}{2}\sin \alpha \right)}^{2}}={{\left( \dfrac{a}{2} \right)}^{2}}\left( {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha \right) \\
\end{align}\]
We know ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$ therefore,
\[{{\left( x-\dfrac{a}{2}\cos \alpha \right)}^{2}}+{{\left( y-\dfrac{a}{2}\sin \alpha \right)}^{2}}={{\left( \dfrac{a}{2} \right)}^{2}}\]
As we know general equation of circle is given by ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ where (h, k) gives center of circle and r gives radius of circle. Comparing above equation with ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ we get that
\[\text{Center}=\left( \dfrac{a}{2}\cos \alpha ,\dfrac{a}{2}\sin \alpha \right)\text{ and Radius}=\dfrac{a}{2}\]
Using same procedure on second equation, consider second equation $r=a\sin \left( \theta -\alpha \right)$
Using $\sin \left( \theta -\alpha \right)=\sin \theta \cos \alpha -\cos \theta \sin \alpha $ we get $r=a\sin \theta \cos \alpha +a\cos \theta \sin \alpha $ converting to Cartesian coordinate by substituting $x=r\cos \theta \text{ and }y=r\sin \theta $ we get:
\[\cos \theta =\dfrac{x}{r}\text{ and }\sin \theta =\dfrac{y}{r}\]
Putting them in above equation, we get:
\[\begin{align}
& r=\dfrac{a}{r}y\cos \alpha -\dfrac{a}{r}x\sin \alpha \\
& \Rightarrow {{r}^{2}}=ay\cos \alpha -ax\sin \alpha \\
\end{align}\]
Using ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$ in above equation, we get:
\[\begin{align}
& {{x}^{2}}+{{y}^{2}}=ay\cos \alpha -ax\sin \alpha \\
& \Rightarrow {{x}^{2}}-ay\cos \alpha +{{y}^{2}}+ax\sin \alpha =0 \\
\end{align}\]
Completing squares of x and y by adding and subtracting ${{\left( \dfrac{a}{2} \right)}^{2}}{{\cos }^{2}}\alpha \text{ and }{{\left( \dfrac{a}{2} \right)}^{2}}{{\sin }^{2}}\alpha $ we get:
\[\begin{align}
& {{x}^{2}}-ay\cos \alpha +{{\left( \dfrac{a}{2} \right)}^{2}}{{\cos }^{2}}\alpha -{{\left( \dfrac{a}{2} \right)}^{2}}{{\cos }^{2}}\alpha +{{y}^{2}}-ax\sin \alpha +{{\left( \dfrac{a}{2} \right)}^{2}}{{\sin }^{2}}\alpha -{{\left( \dfrac{a}{2} \right)}^{2}}{{\sin }^{2}}\alpha =0 \\
& \Rightarrow {{\left( x+\dfrac{a}{2}\sin \alpha \right)}^{2}}+{{\left( y-\dfrac{a}{2}\cos \alpha \right)}^{2}}={{\left( \dfrac{a}{2} \right)}^{2}} \\
\end{align}\]
Comparing above equation by general equation of circle ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ we find that center of above circle is $\left( -\dfrac{a}{2}\sin \alpha ,\dfrac{a}{2}\cos \alpha \right)$ and radius is $\dfrac{a}{2}$.
Hence, we have to find center and radius of both circles. Now we will apply condition to check if both circles cut at right angle which is
\[{{\left( \text{Distance between centers} \right)}^{2}}=\text{Radius}_{1}^{2}+\text{Radius}_{2}^{2}\]
Distance between two points $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ is given by:
\[\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}\]
Applying formula, distance between centers $\left( \dfrac{a}{2}\cos \alpha ,\dfrac{a}{2}\sin \alpha \right)\text{ and }\left( -\dfrac{a}{2}\sin \alpha ,\dfrac{a}{2}\cos \alpha \right)$ is given by
\[\begin{align}
& \sqrt{{{\left( \dfrac{a}{2}\cos \alpha ,\dfrac{a}{2}\sin \alpha \right)}^{2}}+{{\left( -\dfrac{a}{2}\sin \alpha ,\dfrac{a}{2}\cos \alpha \right)}^{2}}} \\
& \Rightarrow \sqrt{\dfrac{{{a}^{2}}}{4}{{\cos }^{2}}\alpha +\dfrac{{{a}^{2}}}{4}{{\sin }^{2}}\alpha +\dfrac{{{a}^{2}}}{2}\sin \alpha \cos \alpha +\dfrac{{{a}^{2}}}{4}{{\sin }^{2}}\alpha +\dfrac{{{a}^{2}}}{4}{{\cos }^{2}}\alpha -\dfrac{{{a}^{2}}}{2}\sin \alpha \cos \alpha } \\
\end{align}\]
Using ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$ we get
\[\begin{align}
& \sqrt{\dfrac{{{a}^{2}}}{4}+\dfrac{{{a}^{2}}}{2}\sin \alpha \cos \alpha +\dfrac{{{a}^{2}}}{4}-\dfrac{{{a}^{2}}}{2}\sin \alpha \cos \alpha } \\
& \Rightarrow \sqrt{\dfrac{{{a}^{2}}}{2}} \\
\end{align}\]
Radius of first circle \[\Rightarrow \text{Radiu}{{\text{s}}_{\text{1}}}=\dfrac{a}{2}\]
Radius of second circle \[\Rightarrow \text{Radiu}{{\text{s}}_{\text{2}}}=\dfrac{a}{2}\]
Putting all of them in ${{\left( \text{Distance between centers} \right)}^{2}}=\text{Radius}_{1}^{2}+\text{Radius}_{2}^{2}$ we get:
\[\begin{align}
& {{\left( \sqrt{\dfrac{{{a}^{2}}}{2}} \right)}^{2}}={{\left( \dfrac{a}{2} \right)}^{2}}+{{\left( \dfrac{a}{2} \right)}^{2}} \\
& \Rightarrow \dfrac{{{a}^{2}}}{2}=\dfrac{{{a}^{2}}}{4}+\dfrac{{{a}^{2}}}{2} \\
\end{align}\]
Therefore, $\dfrac{{{a}^{2}}}{2}=\dfrac{{{a}^{2}}}{2}$
Hence, two circles cut each other at right angle.
Note:
Students should take care while converting polar coordinates to Cartesian coordinates. Students should also learn these formulas and conditions and apply them carefully while solving these types of sums.
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