Question

Prove that the perpendicular at the point of contact of a tangent to a circle passes through the center.

Answer 1

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. Find the equation of line perpendicular to the tangent at the point of contact using the fact that the product of two perpendicular lines is equal to $-1$ . Then show that the equation of the perpendicular has no constant term and hence, will pass through the origin, which is the center of the considered circle.

Complete step by step solution:
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 the 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: Students generally get confused and write the equation of line with slope $m$ and passing through $({{x}_{1}},{{y}_{1}})$ as $(y+{{y}_{1}})=m(x+{{x}_{1}})$ which is wrong. The correct equation of the line should be $(y-{{y}_{1}})=m(x-{{x}_{1}})$ . Such confusions lead students to a wrong answer and hence it should be avoided. Also, the parameter $\theta $ is the angle made by the line joining the origin to the point $(r\cos \theta ,r\sin \theta )$ with positive x axis in the anti-clockwise direction. This point should be very clear as students may take the angle in clockwise direction which is wrong.