Question

A variable line moves in such a way that product of perpendiculars from (a, 0) & (0, 0) is equal to k². The locus of feet of the perpendicular from origin (0, 0) upon the variable line is a circle of radius (\ |a| < 2|k|)

• A. $\sqrt{\frac{a^{2}}{4} + k^{2}}$
• B. $\sqrt{\frac{a^{2}}{4} - k^{2}}$
• C. $\sqrt{a^{2} + \frac{k^{2}}{4}}$
• D. $\sqrt{\frac{a^{2} + k^{2}}{2}}$

Answer 1

Answer: (A)

$\sqrt{\frac{a^{2}}{4} + k^{2}}$

Answer 2

Let line is x cos a + y sin a = p

or xx₁ + yy₁ – (x₁² + y₁²) = 0

[Q cos a = $\frac{x_{1}}{\sqrt{x_{1}^{2} + y_{1}^{2}}}$, sin a = $\frac{y_{1}}{\sqrt{x_{1}^{2} + y_{1}^{2}}}$]

\ ^^r distance OM = $\left| \frac{x_{1}^{2} + y_{1}^{2}}{\sqrt{x_{1}^{2} + y_{1}^{2}}} \right|$ = |$\sqrt{x_{1}^{2} + y_{1}^{2}}$|

& ^^r distance AN = $\frac{|ax_{1} - (x_{1}^{2} + y_{1}^{2})|}{\sqrt{x_{1}^{2} + y_{1}^{2}}}$

\ |$\sqrt{x_{1}^{2} + y_{1}^{2}}$| $\left| \frac{ax_{1} - (x_{1}^{2} + y_{1}^{2})}{\sqrt{x_{1}^{2} + y_{1}^{2}}} \right|$ = k²

Ž |x₁² + y₁² – ax₁| = k²

Ž x₁² + y₁² – ax₁ ± k² = 0

Taking positive sign, radius = $\sqrt{\frac{a^{2}}{4} - k^{2}}$

= Negative (not true)

Taking negative sign, radius = $\sqrt{\frac{a^{2}}{4} + k^{2}}$