# Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Last updated date: 23rd Mar 2023

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Hint: Consider any general equation of circle and take any point on the circle in parametric form. Find the slope of the tangent at the point of contact of the tangent and the circle and the line joining the centre of the circle to the point of contact. Use the fact that the product of slopes of two perpendicular lines is \[-1\].

Take any general equation of circle of the form \[{{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}\]. We have to prove that tangent at any point on the circle is perpendicular to the radius through the point of contact.

We know that the centre of the circle of the form \[{{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}\] is \[O\left( h,k \right)\].

We will consider any general point on the circle of the form \[{{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}\] as \[P\left( h+r\cos \theta ,k+r\sin \theta \right)\].

We will now find the slope of line joining centre \[O\left( h,k \right)\] and \[P\left( h+r\cos \theta ,k+r\sin \theta \right)\].

We know that the slope of line joining two points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] is \[\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\].

Substituting \[{{x}_{1}}=h,{{y}_{1}}=k,{{x}_{2}}=h+r\cos \theta ,{{y}_{2}}=k+r\sin \theta \] in the above equation, we have \[\dfrac{k+r\sin \theta -k}{h+r\cos \theta -h}\] as the slope of line joining centre \[O\left( h,k \right)\] and \[P\left( h+r\cos \theta ,k+r\sin \theta \right)\].

Thus, we have \[{{m}_{1}}=\dfrac{k+r\sin \theta -k}{h+r\cos \theta -h}=\dfrac{r\sin \theta }{r\cos \theta }=\tan \theta \] \[...\left( 1 \right)\] as the slope of line joining centre\[O\left( h,k \right)\] and \[P\left( h+r\cos \theta ,k+r\sin \theta \right)\].

Now, we will find the slope of the tangent at point \[P\left( h+r\cos \theta ,k+r\sin \theta \right)\].

We know that the slope of tangent at any point is represented by \[\dfrac{dy}{dx}\]at that point.

Thus, to find the slope of tangent at point \[P\left( h+r\cos \theta ,k+r\sin \theta \right)\], we will differentiate the equation of circle \[{{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}\].

\[\Rightarrow \dfrac{d}{dx}\left( {{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}} \right)=\dfrac{d}{dx}\left( {{r}^{2}} \right)\]

We will use sum rule of differentiation of functions which states that if \[y=f\left( x \right)+g\left( x \right)\]then\[\dfrac{dy}{dx}=\dfrac{d}{dx}f\left( x \right)+\dfrac{d}{dx}g\left( x \right)\].

Thus, we have \[\dfrac{d}{dx}{{\left( x-h \right)}^{2}}+\dfrac{d}{dx}{{\left( y-k \right)}^{2}}=\dfrac{d}{dx}\left( {{r}^{2}} \right)\]. \[...\left( 2 \right)\]

We know that differentiation of constant is zero with respect to any variable.

We know that differentiation of any function of the form \[y=a{{\left( x-b \right)}^{n}}+c\] is \[\dfrac{dy}{dx}=an{{\left( x-b \right)}^{n-1}}\].

Thus, we have \[\dfrac{d}{dx}{{\left( x-h \right)}^{2}}=2\left( x-h \right)\]. \[...\left( 3 \right)\]

To find the value of \[\dfrac{d}{dx}{{\left( y-k \right)}^{2}}\], we will multiply and divide the given equation by \[dy\].

Thus, we have \[\dfrac{d}{dx}{{\left( y-k \right)}^{2}}=\dfrac{d}{dy}{{\left( y-k \right)}^{2}}\times \dfrac{dy}{dx}\].

We know that \[\dfrac{d}{dy}{{\left( y-k \right)}^{2}}=2\left( y-k \right)\].

Thus, we have \[\dfrac{d}{dx}{{\left( y-k \right)}^{2}}=\dfrac{d}{dy}{{\left( y-k \right)}^{2}}\times \dfrac{dy}{dx}=2\left( y-k \right)\dfrac{dy}{dx}\]. \[...\left( 4 \right)\]

Substituting equation \[\left( 3 \right),\left( 4 \right)\]in equation\[\left( 2 \right)\], we have \[\dfrac{d}{dx}{{\left( x-h \right)}^{2}}+\dfrac{d}{dx}{{\left( y-k \right)}^{2}}=\dfrac{d}{dx}\left( {{r}^{2}} \right)\Rightarrow 2\left( x-h \right)+2\left( y-k \right)\dfrac{dy}{dx}=0\].

Simplifying the equation, we get \[\dfrac{dy}{dx}=\dfrac{h-x}{y-k}\].

Substituting the point \[P\left( h+r\cos \theta ,k+r\sin \theta \right)\] in the above slope of tangent, we have \[\dfrac{dy}{dx}=\dfrac{h-x}{y-k}=\dfrac{h-h-r\cos \theta }{k+r\sin \theta -k}=\dfrac{-r\cos \theta }{r\sin \theta }=-\cot \theta ={{m}_{2}}\] \[...\left( 5 \right)\]

Multiplying equation \[\left( 1 \right)\]and\[\left( 5 \right)\], we get \[{{m}_{1}}{{m}_{2}}=\tan \theta \left( -\cot \theta \right)=-1\].

We know that the product of slopes of two perpendicular lines is \[-1\].

Hence, the line joining the centre of the circle to the point of contact is perpendicular to the tangent of the circle.

Note: We can also take the equation of circle as \[{{x}^{2}}+{{y}^{2}}={{r}^{2}}\] and take any general point on the circle as \[\left( r\cos \theta ,r\sin \theta \right)\]and then find the slope of line joining centre of the circle to the point of contact and the slope of tangent to prove that the two lines are perpendicular to each other.

Take any general equation of circle of the form \[{{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}\]. We have to prove that tangent at any point on the circle is perpendicular to the radius through the point of contact.

We know that the centre of the circle of the form \[{{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}\] is \[O\left( h,k \right)\].

We will consider any general point on the circle of the form \[{{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}\] as \[P\left( h+r\cos \theta ,k+r\sin \theta \right)\].

We will now find the slope of line joining centre \[O\left( h,k \right)\] and \[P\left( h+r\cos \theta ,k+r\sin \theta \right)\].

We know that the slope of line joining two points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] is \[\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\].

Substituting \[{{x}_{1}}=h,{{y}_{1}}=k,{{x}_{2}}=h+r\cos \theta ,{{y}_{2}}=k+r\sin \theta \] in the above equation, we have \[\dfrac{k+r\sin \theta -k}{h+r\cos \theta -h}\] as the slope of line joining centre \[O\left( h,k \right)\] and \[P\left( h+r\cos \theta ,k+r\sin \theta \right)\].

Thus, we have \[{{m}_{1}}=\dfrac{k+r\sin \theta -k}{h+r\cos \theta -h}=\dfrac{r\sin \theta }{r\cos \theta }=\tan \theta \] \[...\left( 1 \right)\] as the slope of line joining centre\[O\left( h,k \right)\] and \[P\left( h+r\cos \theta ,k+r\sin \theta \right)\].

Now, we will find the slope of the tangent at point \[P\left( h+r\cos \theta ,k+r\sin \theta \right)\].

We know that the slope of tangent at any point is represented by \[\dfrac{dy}{dx}\]at that point.

Thus, to find the slope of tangent at point \[P\left( h+r\cos \theta ,k+r\sin \theta \right)\], we will differentiate the equation of circle \[{{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}\].

\[\Rightarrow \dfrac{d}{dx}\left( {{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}} \right)=\dfrac{d}{dx}\left( {{r}^{2}} \right)\]

We will use sum rule of differentiation of functions which states that if \[y=f\left( x \right)+g\left( x \right)\]then\[\dfrac{dy}{dx}=\dfrac{d}{dx}f\left( x \right)+\dfrac{d}{dx}g\left( x \right)\].

Thus, we have \[\dfrac{d}{dx}{{\left( x-h \right)}^{2}}+\dfrac{d}{dx}{{\left( y-k \right)}^{2}}=\dfrac{d}{dx}\left( {{r}^{2}} \right)\]. \[...\left( 2 \right)\]

We know that differentiation of constant is zero with respect to any variable.

We know that differentiation of any function of the form \[y=a{{\left( x-b \right)}^{n}}+c\] is \[\dfrac{dy}{dx}=an{{\left( x-b \right)}^{n-1}}\].

Thus, we have \[\dfrac{d}{dx}{{\left( x-h \right)}^{2}}=2\left( x-h \right)\]. \[...\left( 3 \right)\]

To find the value of \[\dfrac{d}{dx}{{\left( y-k \right)}^{2}}\], we will multiply and divide the given equation by \[dy\].

Thus, we have \[\dfrac{d}{dx}{{\left( y-k \right)}^{2}}=\dfrac{d}{dy}{{\left( y-k \right)}^{2}}\times \dfrac{dy}{dx}\].

We know that \[\dfrac{d}{dy}{{\left( y-k \right)}^{2}}=2\left( y-k \right)\].

Thus, we have \[\dfrac{d}{dx}{{\left( y-k \right)}^{2}}=\dfrac{d}{dy}{{\left( y-k \right)}^{2}}\times \dfrac{dy}{dx}=2\left( y-k \right)\dfrac{dy}{dx}\]. \[...\left( 4 \right)\]

Substituting equation \[\left( 3 \right),\left( 4 \right)\]in equation\[\left( 2 \right)\], we have \[\dfrac{d}{dx}{{\left( x-h \right)}^{2}}+\dfrac{d}{dx}{{\left( y-k \right)}^{2}}=\dfrac{d}{dx}\left( {{r}^{2}} \right)\Rightarrow 2\left( x-h \right)+2\left( y-k \right)\dfrac{dy}{dx}=0\].

Simplifying the equation, we get \[\dfrac{dy}{dx}=\dfrac{h-x}{y-k}\].

Substituting the point \[P\left( h+r\cos \theta ,k+r\sin \theta \right)\] in the above slope of tangent, we have \[\dfrac{dy}{dx}=\dfrac{h-x}{y-k}=\dfrac{h-h-r\cos \theta }{k+r\sin \theta -k}=\dfrac{-r\cos \theta }{r\sin \theta }=-\cot \theta ={{m}_{2}}\] \[...\left( 5 \right)\]

Multiplying equation \[\left( 1 \right)\]and\[\left( 5 \right)\], we get \[{{m}_{1}}{{m}_{2}}=\tan \theta \left( -\cot \theta \right)=-1\].

We know that the product of slopes of two perpendicular lines is \[-1\].

Hence, the line joining the centre of the circle to the point of contact is perpendicular to the tangent of the circle.

Note: We can also take the equation of circle as \[{{x}^{2}}+{{y}^{2}}={{r}^{2}}\] and take any general point on the circle as \[\left( r\cos \theta ,r\sin \theta \right)\]and then find the slope of line joining centre of the circle to the point of contact and the slope of tangent to prove that the two lines are perpendicular to each other.

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