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Question Number 77131 by peter frank last updated on 03/Jan/20

	Answered by mr W last updated on 03/Jan/20

	$p e r p e n d i c u l a r t a n g e n t s f r o m P (u, v) :$ $y = v + m (x - u) \Rightarrow m x - y + (v - m u)$ $y = v - \frac{1}{m} (x - u) \Rightarrow x + m y - (m v + u)$ $f r o m Q 77127 w e h a v e :$ $a^{2} m^{2} + b^{2} = {(v - m u)}^{2}$ $\Rightarrow a^{2} m^{2} + b^{2} = v^{2} - 2 m u v + m^{2} u^{2} . . . (i)$ $a^{2} + b^{2} m^{2} = {(m v + u)}^{2}$ $\Rightarrow a^{2} + b^{2} m^{2} = m^{2} v^{2} + 2 m u v + u^{2} . . . (i i)$ $(i) + (i i) :$ $(1 + m^{2}) a^{2} + (1 + m^{2}) b^{2} = (1 + m^{2}) v^{2} + (1 + m^{2}) u^{2}$ $\Rightarrow u^{2} + v^{2} = a^{2} + b^{2}$ $i . e . l o c u s o f p o i n t P (x, y) i s c i r c l e :$ $\Rightarrow x^{2} + y^{2} = a^{2} + b^{2}$
	Commented by peter frank last updated on 06/Jan/20

	$t h a n k f o r y o u r h e l p . I h a v e$ $l e a r n t s o m u c h s i r$
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