Find-locus-of-point-P-from-which-tangents-PA-amp-PB-to-circles-x-2-y-2-a-2-and-x-2-y-2-b-2-respectively-are-perpendicular-

Question Number 47621 by rahul 19 last updated on 12/Nov/18

F i n d l o c u s o f p o i n t P f r o m w h i c h

t a n g e n t s P A & P B t o c i r c l e s x^{2} + y^{2} = a^{2}

a n d x^{2} + y^{2} = b^{2} r e s p e c t i v e l y a r e p e r p e n d i c u l a r .

Commented by rahul 19 last updated on 12/Nov/18

A n s \to x^{2} + y^{2} = a^{2} + b^{2} .

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Nov/18

t a n g e n t o n x^{2} + y^{2} = a^{2} f r o m p (α, β) i s

y - β = m (x - α)

y - m x + m α - β = 0

d i s t a n c e f r o m (0, 0) i s r a d i u s

∣ \frac{0 - 0 + m α - β}{\sqrt{1^{2} + m^{2}}} ∣= a

\frac{m α - β}{\sqrt{1 + m^{2}}} = a

t a n g e n t o n x^{2} + y^{2} = b^{2}

y - β = - \frac{1}{m} (x - α)

m y - m β + x - α = 0

∣ \frac{- m β - α}{\sqrt{1 + m^{2}}} ∣= b

\frac{m β + α}{\sqrt{1 + m^{2}}} = b

a^{2} + b^{2} = \frac{{(m α - β)}^{2} + {(m β + α)}^{2}}{1 + m^{2}}

a^{2} + b^{2} = \frac{α^{2} (1 + m^{2}) + β^{2} (1 + m^{2})}{(1 + m^{2})}

a^{2} + b^{2} = α^{2} + β^{2}

h e n c e l o c u s i s

x^{2} + y^{2} = a^{2} + b^{2}

Commented by rahul 19 last updated on 13/Nov/18

thanks sir ��