Find the equation of circle in complex form which touches iz + z^� + 1 + i = 0 and for which the lines (1 − i)z = (1 + i)z^� and (1 + i)z + (i − 1)z^� − 4i = 0 are normals.


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 20550 by Tinkutara last updated on 28/Aug/17

$F i n d t h e e q u a t i o n o f c i r c l e i n c o m p l e x$ $f o r m w h i c h t o u c h e s i z + \bar{z} + 1 + i = 0$ $a n d f o r w h i c h t h e l i n e s (1 - i) z =$ $(1 + i) \bar{z} a n d (1 + i) z + (i - 1) \bar{z} - 4 i = 0$ $a r e n o r m a l s .$
	Answered by ajfour last updated on 30/Aug/17

	$C e n t e r o f c i r c l e l i e s o n b o t h t h e$ $n o r m a l s : l e t c e n t e r b e z_{0} .$ $\Rightarrow (1 + i) z_{0} - (1 - i) {\bar{z}}_{0} = 4 i$ $(1 - i) z_{0} - (1 + i) {\bar{z}}_{0} = 0$ $a d d i n g : 2 z_{0} - 2 {\bar{z}}_{0} = 4 i$ $o r z_{0} - {\bar{z}}_{0} = 2 i . . . . (i)$ $s u b t r a c t i n g : 2 {i z}_{0} + 2 i {\bar{z}}_{0} = 4 i$ $o r z_{0} + {\bar{z}}_{0} = 2 . . . . (i i)$ $a d d i n g (i) a n d (i i) :$ $z_{0} = 1 + i$ $H e n c e e q u a t i o n o f c i r c l e c a n b e$ $a s s u m e d t o b e :$ $∣ z - z_{0} ∣= r a n d l e t g i v e n t a n g e n t$ $t o c i r c l e t o u c h e s i t a t z_{1} .$ $T h e n ∣ z_{1} - z_{0} ∣= r w h e r e (z_{0} = 1 + i)$ $o r (z_{1} - z_{0}) ({\bar{z}}_{1} - {\bar{z}}_{0}) = r^{2} . . . (i i i)$ $f r o m e q u a t i o n o f t a n g e n t$ ${i z}_{1} + {\bar{z}}_{1} + 1 + i = 0$ $\Rightarrow {\bar{z}}_{1} = - ({i z}_{1} + 1 + i)$ $s u b s t i t u t i n g i n (i i i) {w i t h z}_{0} = 1 + i$ $(z_{1} - 1 - i) (- {i z}_{1} - 1 - i - 1 + i) = r^{2}$ $(z_{1} - 1 - i) ({i z}_{1} + 1 + i + 1 - i) = - r^{2}$ $\Rightarrow (z_{1} - 1 - i) ({i z}_{1} + 2) = - r^{2}$ $o r {i z}_{1}^{2} + 2 z_{1} - {i z}_{1} - 2 + z_{1} - 2 i + r^{2} = 0$ ${i z}_{1}^{2} + (3 - i) z_{1} - (2 - r^{2} + 2 i) = 0$ $a s z_{1} i s a d o u b l e r o o t (l i n e i s$ $t a n g e n t t o c i r c l e) w e m u s t h a v e$ $D = 0, i m p l i e s$ ${(3 - i)}^{2} + 4 i (2 - r^{2} + 2 i) = 0$ $\Rightarrow 8 - 6 i + 8 i - 4 {i r}^{2} - 8 = 0$ $\Rightarrow 4 r^{2} = 2 o r r = \frac{1}{\sqrt{2}} .$ $H e n c e e q u a t i o n o f c i r c l e i s$ $∣ z - 1 - i ∣= \frac{1}{\sqrt{2}}$ $o r (z - 1 - i) (\bar{z} - 1 + i) = \frac{1}{2}$ $z \bar{z} + (i - 1) z - (1 + i) \bar{z} + 2 - \frac{1}{2} = 0$ $z \bar{z} + (i - 1) z - (1 + i) \bar{z} + 3 / 2 = 0 .$
	Commented by Tinkutara last updated on 30/Aug/17

	$Thank you very much Sir!$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum