Find locus of point P from which tangents PA & PB to circles x^2 +y^2 =a^2 and x^2 +y^2 =b^2 respectively are perpendicular.


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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 byrahul 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 byrahul 19 last updated on 13/Nov/18
	thanks sir ��
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