Question

: Find the value of p so that the straight line \[x\cos \alpha +y\sin \alpha -p=0\] may touch the circle \[{{x}^{2}}+{{y}^{2}}-2ax\cos \alpha -2b\sin \alpha -{{a}^{2}}{{\sin }^{2}}\alpha =0\].

Answer 1

Hint: First, let the straight-line equation touch the circle at point p. Now, we let the coordinates of the p be $({{x}_{1}},{{y}_{1}})$. For obtaining the values of x1 and the y1 respectively, that is the coordinates of the point p two equations are made from the problem statement.

Complete step by step answer:
According to the question, we are given a straight line \[x\cos \alpha +y\sin \alpha -p=0\]. Let, the straight line \[x\cos \alpha +y\sin \alpha =p\] touches the curve at the point P.
Let the coordinates of P be ${{x}_{1}}\text{ }and\text{ }{{\text{y}}_{1}}$ respectively.
Now, the equation of the tangent at the point P (x1, y1) to any circle ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$ is given as:
$\dfrac{x{{x}_{1}}}{{{a}^{2}}}-\dfrac{y{{y}_{1}}}{{{b}^{2}}}=1...(1)$
Consider the above equation number as the equation (1).
Now, we take the equation $x\cos \alpha +y\sin \alpha =p$ because both the equation represents the same line as per the problem statement, since tangent also touches the circle at one point.
$\dfrac{x}{a}\cos \alpha +y\dfrac{\sin \alpha }{p}=1...(2)$
Consider the above equation number as the equation (2).
Now, as you can see that the coefficient of both the equations (1) and (2) are the same. Now, comparing each coefficient to obtain the individual value of coordinates:
$\begin{align}
  & \dfrac{{{x}_{1}}}{{{\alpha }^{2}}}=\dfrac{\cos \alpha }{p} \\
 & -\dfrac{{{y}_{1}}}{{{b}^{2}}}=\dfrac{\sin \alpha }{p} \\
\end{align}$
So, the ${{x}_{1}}=\dfrac{{{a}^{2}}cos\alpha }{p}\text{ and }{{\text{y}}_{1}}=\dfrac{{{b}^{2}}cos\alpha }{p}$
These points will always lie on the given equation of the straight line. So, satisfying them in the straight line, we get
\[\begin{align}
  & \therefore \dfrac{{{a}^{2}}{{\cos }^{2}}\alpha }{p}-\dfrac{{{b}^{2}}{{\sin }^{2}}\alpha }{p}=p \\
 & {{a}^{2}}{{\cos }^{2}}\alpha -{{b}^{2}}{{\sin }^{2}}\alpha ={{p}^{2}} \\
\end{align}\]
Hence, it proved that \[{{a}^{2}}{{\cos }^{2}}\alpha -{{b}^{2}}{{\sin }^{2}}\alpha ={{p}^{2}}\]. So, we obtained the desired result.

Note: The key step for solving this problem is the knowledge of tangent was circle in the general form. By using the equation of tangent and comparing it with the given equation, students will get the answer without any error.