Question

The coordinate that the chord \[x\cos \alpha + y\sin \alpha - p = 0\] of \[{x^2} + {y^2} - {a^2} = 0\] may subtend a right angle at the center of the circle is?
A) \[{a^2} = 2{p^2}\]
B) \[{p^2} = 2{a^2}\]
C) \[a = 2p\]
D) \[p = 2a\]

Answer 1

Hint: first we compare the given equation with the standard equation of circle so we get centre of a circle is origin then we write the homogeneous equation of second degree and after simplifying that we will get the condition.

Complete step by step solution: The combined equation of the lines joining the origin to the points of intersection of \[x\cos \alpha + y\sin \alpha - p = 0\]and \[{x^2} + {y^2} - {a^2} = 0\] is homogeneous equation of second degree given by
\[{x^2} + {y^2} - {a^2}{\left( {\dfrac{{x\cos \alpha + y\sin \alpha }}{p}} \right)^2} = 0\]
\[ \Rightarrow \]\[\left[ {{x^2}\left( {{p^2} - {a^2}{{\cos }^2}\alpha } \right) + {y^2}\left( {{p^2} - {a^2}{{\sin }^2}\alpha } \right) - 2xy{a^2}\sin \alpha \cos \alpha } \right] = 0\]
The lines given by this equation are at right angle if
\[\left( {{p^2} - {a^2}{{\cos }^2}\alpha } \right)\]+\[\left( {{p^2} - {a^2}{{\sin }^2}\alpha } \right)\]\[ = 0\]
\[ \Rightarrow 2{p^2} - {a^2}\left( {{{\cos }^2}\alpha + {{\sin }^2}\alpha } \right) = 0\]
\[ \Rightarrow 2{p^2} = {a^2}\left( {{{\cos }^2}\alpha + {{\sin }^2}\alpha } \right)\]
\[ \Rightarrow 2{p^2} = {a^2}\]
\[ \Rightarrow {a^2} = 2{p^2}\]
Hence, option A. \[{a^2} = 2{p^2}\] is the correct answer.


Note: there is an alternative solution to this question which is given as follows.
The distance of the given line from the circle of the circle is \[\left| {\left. p \right|} \right.\]
Now, the line subtends a right angle at the centre.
Hence, radius\[ = \sqrt 2 \left| {\left. p \right|} \right.\]
Or a\[ = \sqrt 2 \left| {\left. p \right|} \right.\]
 \[ \Rightarrow {a^2} = 2{p^2}\]