Question

The locus of the feet of perpendiculars drawn from the point $(a,0)$ on tangents to the circle $x^2+y^2=a^2$ is

• A. $a^2(x^2+y^2+ax{)}^2=a^2(y^2+(x+a{)}^2)$
• B. $a^2(x^2+y^2-ax{)}^2=y^2+(x-a{)}^2$
• C. $(x^2+y^2-ax{)}^2=a^2(y^2+(x-a{)}^2)$
• D. $a^2\left[\right(x^2+y^2)-a^2{x}^2]=(y^2+(x-a{)}^2)$

Answer 1

The correct option is C $(x^2+y^2-ax{)}^2=a^2(y^2+(x-a{)}^2)$

$x\cos \theta +y\sin \theta =a$ is equation of tangent to $x^2+y^2=a^2$
Let P(h,k) is feet of perpendicular from (a,o) to $x\cos \theta +y\sin \theta =a$
$\frac{x-a}{\cos \theta }=\frac{y-o}{\sin \theta }=-\frac{\left(a\right(\cos \theta -1\left)\right)}{1}$
$x=a-a\cos \theta (\cos \theta -1)andy=-a\sin \theta (\cos \theta -1)$
$\therefore (x-a{)}^2+y^2=a^2(\cos \theta -1{)}^2\times \left(1\right)$ ---(1)

$=-a^2\cos \theta (\cos \theta -1)+a^2(\cos \theta -1{)}^2$
$a^2(\cos \theta -1)\left[\right(\cos \theta -1)-\cos \theta ]$
$(x-a{)}^2+y^2=a^2(1-\cos \theta )$ --- (2)
$\therefore$ from (1) & (2),
$\Rightarrow (x^2+y^2-ax{)}^2=a^2(y^2+(x-a{)}^2)$