Question

Find the value of p, so that the straight line $x\cos \alpha + y\sin \alpha = p$ may touch the circle ${x^2} + {y^2} - 2ax\cos \alpha - 2ay\sin \alpha = 0$.

Answer 1

Hint- Here, we know that since the given straight line is tangent of the given circle so the perpendicular distance of the centre of the given circle from the given straight line will be equal to the radius of the circle and this perpendicular distance is calculated using the formula $d = \left| {\dfrac{{A{x_1} + B{y_1} + c}}{{\sqrt {{A^2} + {B^2}} }}} \right|$.

Complete step-by-step solution -
Given, equation of straight line is $x\cos \alpha + y\sin \alpha = p \Rightarrow x\cos \alpha + y\sin \alpha - p = 0$
It is also given that the above straight line touches the circle whose equation is ${x^2} + {y^2} - 2ax\cos \alpha - 2ay\sin \alpha = 0$
For the given straight line to touch the given circle, that straight line must be a tangent to the circle.
Also for the given straight line to be the tangent of the given circle, the perpendicular distance of the centre of the circle from the given straight line should be equal to the radius of the given circle.
As we know that the equation of any circle with centre coordinate as $\left( {{x_1},{y_1}} \right)$ and radius of the circle as r is given by ${\left( {x - {x_1}} \right)^2} + {\left( {y - {y_1}} \right)^2} = {r^2}{\text{ }} \to {\text{(1)}}$
Now, let us convert the given equation of the circle in terms of the above equation by completing the square method.
\[
  {x^2} + {y^2} - 2ax\cos \alpha - 2ay\sin \alpha = 0 \Rightarrow {x^2} - 2ax\cos \alpha + {\left( {a\cos \alpha } \right)^2} + {y^2} - 2ay\sin \alpha + {\left( {a\sin \alpha } \right)^2} - {\left( {a\cos \alpha } \right)^2} - {\left( {a\sin \alpha } \right)^2} = 0 \\
   \Rightarrow {\left( {x - a\cos \alpha } \right)^2} + {\left( {y - a\sin \alpha } \right)^2} = {\left( {a\cos \alpha } \right)^2} + {\left( {a\sin \alpha } \right)^2} \\
   \Rightarrow {\left( {x - a\cos \alpha } \right)^2} + {\left( {y - a\sin \alpha } \right)^2} = {a^2}{\left( {\cos \alpha } \right)^2} + {a^2}{\left( {\sin \alpha } \right)^2} \\
   \Rightarrow {\left( {x - a\cos \alpha } \right)^2} + {\left( {y - a\sin \alpha } \right)^2} = {a^2}\left[ {{{\left( {\cos \alpha } \right)}^2} + {{\left( {\sin \alpha } \right)}^2}} \right] \\
 \]
Using the formula, \[{\left( {\cos \alpha } \right)^2} + {\left( {\sin \alpha } \right)^2} = 1\] , we get
\[ \Rightarrow {\left( {x - a\cos \alpha } \right)^2} + {\left( {y - a\sin \alpha } \right)^2} = {a^2}\]
Comparing the above equation of given circle obtained with the general form of any circle given by equation (1), we can write
The centre coordinate of the given circle is C$\left( {a\cos \alpha ,a\sin \alpha } \right)$ and the radius is a.
Also, we know that the distance of any point P$\left( {{x_1},{y_1}} \right)$ from a straight line having equation $Ax + By + c = 0$ is given by
$d = \left| {\dfrac{{A{x_1} + B{y_1} + c}}{{\sqrt {{A^2} + {B^2}} }}} \right|$
Using above formula, we can find the value of p
Therefore, distance of the centre C$\left( {a\cos \alpha ,a\sin \alpha } \right)$ from the straight line$x\cos \alpha + y\sin \alpha - p = 0$ will be
 $
  d = \left| {\dfrac{{\left( {a\cos \alpha } \right)\cos \alpha + \left( {a\sin \alpha } \right)\sin \alpha - p}}{{\sqrt {{{\left( {\cos \alpha } \right)}^2} + {{\left( {\sin \alpha } \right)}^2}} }}} \right| = \left| {\dfrac{{a{{\left( {\cos \alpha } \right)}^2} + a{{\left( {\sin \alpha } \right)}^2} - p}}{{\sqrt 1 }}} \right| \\
   \Rightarrow d = \left| {a\left[ {{{\left( {\cos \alpha } \right)}^2} + {{\left( {\sin \alpha } \right)}^2}} \right] - p} \right| = \left| {a - p} \right| \\
 $
Either $d = a - p$ or $
  d = - \left( {a - p} \right) \\
   \Rightarrow d = - a + p \\
 $
Since, $d = r = a$
Either $
  a = a - p \\
   \Rightarrow p = 0 \\
 $ or $
  a = - a + p \\
   \Rightarrow p = 2a \\
 $
Therefore, the required values of p can be 0 or 2a.

Note- In these types of problems, we convert the given equations of the straight line and the circle in the same form as required by the formulas which we have used in the solution. Here, it is very important to note that if a straight line touches any circle then that straight line must be tangent to the circle.