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

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$= \sqrt 2 \left| {\left. p \right|} \right.$
$\Rightarrow {a^2} = 2{p^2}$