Question

Locus of the points from which perpendicular tangent can be drawn to the circle $x ^ { 2 } + y ^ { 2 } = a ^ { 2 }$ , is.

• A. A circle passing through origin
• B. A circle of radius 2a
• C. A concentric circle of radius $a \sqrt { 2 }$
• D. None of these

Answer 1

Answer: (C)

A concentric circle of radius $a \sqrt { 2 }$

Answer 2

Required locus is $S S _ { 1 } = T ^ { 2 }$

$\left( x ^ { 2 } + y ^ { 2 } - a ^ { 2 } \right) \left( h ^ { 2 } + k ^ { 2 } - a ^ { 2 } \right) = \left( h x + k y - a ^ { 2 } \right) ^ { 2 }$

But as given, coefficient of $x ^ { 2 } +$ coefficient of $y ^ { 2 } = 0$

$\Rightarrow$ $h ^ { 2 } + k ^ { 2 } = 2 a ^ { 2 }$ .

Hence locus of the point is the circle with centre (0, 0) and radius $a \sqrt { 2 }$.