Prove that the perpendicular at the point of contact of a tangent to a circle passes through the center.
Last updated date: 23rd Mar 2023
•
Total views: 307.5k
•
Views today: 2.85k
Answer
307.5k+ views
Hint: The given problem is related to the equation of the tangent to a circle. Try to remember the equation of a tangent to a circle in parametric form.
Complete step-by-step answer:
We will consider the circle ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$. We know the radius of this circle is $r$ and the center of this circle is at $(0,0)$ .
We will consider a point $x=r\cos \theta $ and $y=r\sin \theta $ , i.e. $(r\cos \theta ,r\sin \theta )$ , on the circle, where $\theta $ is a parameter.
We know, the equation of the tangent at $({{x}_{1}},{{y}_{1}})$ is given as $x{{x}_{1}}+y{{y}_{1}}={{r}^{2}}$ .
So, the equation of the tangent at $(r\cos \theta ,r\sin \theta )$ is given as $x.r\cos \theta +y.r\sin \theta ={{r}^{2}}$
$\Rightarrow x\cos \theta +y\sin \theta =r....(i)$
Now, we know, the slope of the line given by $ax+by+c=0$ is given as $m=-\dfrac{a}{b}$
So, the slope of tangent given by equation$(i)$ is given as $m=-\dfrac{\cos \theta }{\sin \theta }=-\cot \theta $
Now, we know the product of two perpendicular lines is equal to $-1$ .
Let ${{m}_{\bot }}$ be the slope of the line perpendicular to the tangent. So, $m\times {{m}_{\bot }}=-1$
$\Rightarrow -\cot \theta \times {{m}_{\bot }}=-1$
$\Rightarrow {{m}_{\bot }}=\dfrac{-1}{-\cot \theta }=\tan \theta $
So, the slope of the line perpendicular to the tangent at $(r\cos \theta ,r\sin \theta )$ is given as ${{m}_{\bot }}=\tan \theta $ .
Now, we know, the equation of a line with slope $m$ and passing through $({{x}_{1}},{{y}_{1}})$ is given as $(y-{{y}_{1}})=m(x-{{x}_{1}})$.
So, the equation of the line passing through $(r\cos \theta ,r\sin \theta )$ and having slope ${{m}_{\bot }}=\tan \theta $ is given as $y-r\sin \theta =\tan \theta (x-r\cos \theta )$ .
$\Rightarrow y-r\sin \theta =x\tan \theta -r\sin \theta $
$\Rightarrow y-x\tan \theta =0.....(ii)$
Now, there is no constant term in equation $(ii)$ . So, the line represented by equation $(ii)$ will always pass through the origin, which is the center of the circle ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$.
Hence, any line perpendicular at the point of contact of a tangent to a circle passes through the center of the circle.
Note: While making substitutions, make sure that the substitutions are done correctly and no sign mistakes are present. Sign mistakes can cause the final answer to be wrong.
Complete step-by-step answer:
We will consider the circle ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$. We know the radius of this circle is $r$ and the center of this circle is at $(0,0)$ .
We will consider a point $x=r\cos \theta $ and $y=r\sin \theta $ , i.e. $(r\cos \theta ,r\sin \theta )$ , on the circle, where $\theta $ is a parameter.
We know, the equation of the tangent at $({{x}_{1}},{{y}_{1}})$ is given as $x{{x}_{1}}+y{{y}_{1}}={{r}^{2}}$ .
So, the equation of the tangent at $(r\cos \theta ,r\sin \theta )$ is given as $x.r\cos \theta +y.r\sin \theta ={{r}^{2}}$
$\Rightarrow x\cos \theta +y\sin \theta =r....(i)$
Now, we know, the slope of the line given by $ax+by+c=0$ is given as $m=-\dfrac{a}{b}$
So, the slope of tangent given by equation$(i)$ is given as $m=-\dfrac{\cos \theta }{\sin \theta }=-\cot \theta $
Now, we know the product of two perpendicular lines is equal to $-1$ .
Let ${{m}_{\bot }}$ be the slope of the line perpendicular to the tangent. So, $m\times {{m}_{\bot }}=-1$
$\Rightarrow -\cot \theta \times {{m}_{\bot }}=-1$
$\Rightarrow {{m}_{\bot }}=\dfrac{-1}{-\cot \theta }=\tan \theta $
So, the slope of the line perpendicular to the tangent at $(r\cos \theta ,r\sin \theta )$ is given as ${{m}_{\bot }}=\tan \theta $ .
Now, we know, the equation of a line with slope $m$ and passing through $({{x}_{1}},{{y}_{1}})$ is given as $(y-{{y}_{1}})=m(x-{{x}_{1}})$.
So, the equation of the line passing through $(r\cos \theta ,r\sin \theta )$ and having slope ${{m}_{\bot }}=\tan \theta $ is given as $y-r\sin \theta =\tan \theta (x-r\cos \theta )$ .
$\Rightarrow y-r\sin \theta =x\tan \theta -r\sin \theta $
$\Rightarrow y-x\tan \theta =0.....(ii)$
Now, there is no constant term in equation $(ii)$ . So, the line represented by equation $(ii)$ will always pass through the origin, which is the center of the circle ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$.

Hence, any line perpendicular at the point of contact of a tangent to a circle passes through the center of the circle.
Note: While making substitutions, make sure that the substitutions are done correctly and no sign mistakes are present. Sign mistakes can cause the final answer to be wrong.
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
