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Question Number 20550 by Tinkutara last updated on 28/Aug/17

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.

$${Find}\:{the}\:{equation}\:{of}\:{circle}\:{in}\:{complex} \\ $$$${form}\:{which}\:{touches}\:{iz}\:+\:\bar {{z}}\:+\:\mathrm{1}\:+\:{i}\:=\:\mathrm{0} \\ $$$${and}\:{for}\:{which}\:{the}\:{lines}\:\left(\mathrm{1}\:−\:{i}\right){z}\:= \\ $$$$\left(\mathrm{1}\:+\:{i}\right)\bar {{z}}\:{and}\:\left(\mathrm{1}\:+\:{i}\right){z}\:+\:\left({i}\:−\:\mathrm{1}\right)\bar {{z}}\:−\:\mathrm{4}{i}\:=\:\mathrm{0} \\ $$$${are}\:{normals}. \\ $$

Answered by ajfour last updated on 30/Aug/17

Center of circle lies on both the normals: let center be z_0 . ⇒ (1+i)z_0 −(1−i)z_0 ^� =4i (1−i)z_0 −(1+i)z_0 ^� =0 adding: 2z_0 −2z_0 ^� =4i or z_0 −z_0 ^� =2i ....(i) subtracting: 2iz_0 +2iz_0 ^� =4i or z_0 +z_0 ^� =2 ....(ii) adding (i) and (ii): z_0 = 1+i Hence equation of circle can be assumed to be: ∣z−z_0 ∣=r and let given tangent to circle touches it at z_1 . Then ∣z_1 −z_0 ∣=r where (z_0 =1+i) or (z_1 −z_0 )(z_1 ^� −z_0 ^� )=r^2 ...(iii) from equation of tangent iz_1 +z_1 ^� +1+i=0 ⇒ z_1 ^� =−(iz_1 +1+i) substituting in (iii) withz_0 =1+i (z_1 −1−i)(−iz_1 −1−i−1+i)=r^2 (z_1 −1−i)(iz_1 +1+i+1−i)=−r^2 ⇒ (z_1 −1−i)(iz_1 +2)=−r^2 or iz_1 ^2 +2z_1 −iz_1 −2+z_1 −2i+r^2 =0 iz_1 ^2 +(3−i)z_1 −(2−r^2 +2i)=0 as z_1 is a double root (line is tangent to circle) we must have D=0 , implies (3−i)^2 +4i(2−r^2 +2i)=0 ⇒ 8−6i+8i−4ir^2 −8=0 ⇒ 4r^2 =2 or r=(1/(√2)) . Hence equation of circle is ∣z−1−i∣=(1/(√2)) or (z−1−i)(z^� −1+i)=(1/2) zz^� +(i−1)z−(1+i)z^� +2−(1/2)=0 zz^� +(i−1)z−(1+i)z^� +3/2=0 .

$${Center}\:{of}\:{circle}\:{lies}\:{on}\:{both}\:{the}\: \\ $$$${normals}:\:{let}\:{center}\:{be}\:{z}_{\mathrm{0}} . \\ $$$$\Rightarrow\:\:\:\left(\mathrm{1}+{i}\right){z}_{\mathrm{0}} −\left(\mathrm{1}−{i}\right)\bar {{z}}_{\mathrm{0}} =\mathrm{4}{i} \\ $$$$\:\:\:\:\:\:\:\left(\mathrm{1}−{i}\right){z}_{\mathrm{0}} −\left(\mathrm{1}+{i}\right)\bar {{z}}_{\mathrm{0}} \:=\mathrm{0} \\ $$$${adding}:\:\:\:\:\:\:\:\mathrm{2}{z}_{\mathrm{0}} −\mathrm{2}\bar {{z}}_{\mathrm{0}} =\mathrm{4}{i} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{or}\:\:\:{z}_{\mathrm{0}} −\bar {{z}}_{\mathrm{0}} =\mathrm{2}{i}\:\:\:\:....\left({i}\right) \\ $$$${subtracting}:\:\:\:\mathrm{2}{iz}_{\mathrm{0}} +\mathrm{2}{i}\bar {{z}}_{\mathrm{0}} =\mathrm{4}{i} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{or}\:\:{z}_{\mathrm{0}} +\bar {{z}}_{\mathrm{0}} =\mathrm{2}\:\:\:\:\:\:....\left({ii}\right) \\ $$$${adding}\:\left({i}\right)\:{and}\:\left({ii}\right): \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{z}}_{\mathrm{0}} \:=\:\mathrm{1}+\boldsymbol{{i}} \\ $$$${Hence}\:{equation}\:{of}\:{circle}\:{can}\:{be} \\ $$$${assumed}\:{to}\:{be}: \\ $$$$\:\:\:\mid{z}−{z}_{\mathrm{0}} \mid={r}\:\:{and}\:{let}\:{given}\:{tangent} \\ $$$${to}\:{circle}\:{touches}\:{it}\:{at}\:{z}_{\mathrm{1}} . \\ $$$${Then}\:\:\:\mid{z}_{\mathrm{1}} −{z}_{\mathrm{0}} \mid={r}\:\:{where}\:\left({z}_{\mathrm{0}} =\mathrm{1}+{i}\right) \\ $$$${or}\:\:\:\:\left({z}_{\mathrm{1}} −{z}_{\mathrm{0}} \right)\left(\bar {{z}}_{\mathrm{1}} −\bar {{z}}_{\mathrm{0}} \right)={r}^{\mathrm{2}} \:\:\:\:...\left({iii}\right) \\ $$$${from}\:{equation}\:{of}\:{tangent} \\ $$$$\:\:\:\:{iz}_{\mathrm{1}} +\bar {{z}}_{\mathrm{1}} +\mathrm{1}+{i}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\bar {{z}}_{\mathrm{1}} =−\left({iz}_{\mathrm{1}} +\mathrm{1}+{i}\right) \\ $$$${substituting}\:{in}\:\left({iii}\right)\:{withz}_{\mathrm{0}} =\mathrm{1}+{i} \\ $$$$\:\left({z}_{\mathrm{1}} −\mathrm{1}−{i}\right)\left(−{iz}_{\mathrm{1}} −\mathrm{1}−{i}−\mathrm{1}+{i}\right)={r}^{\mathrm{2}} \\ $$$$\:\left({z}_{\mathrm{1}} −\mathrm{1}−{i}\right)\left({iz}_{\mathrm{1}} +\mathrm{1}+{i}+\mathrm{1}−{i}\right)=−{r}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\left({z}_{\mathrm{1}} −\mathrm{1}−{i}\right)\left({iz}_{\mathrm{1}} +\mathrm{2}\right)=−{r}^{\mathrm{2}} \\ $$$${or}\:\:\:{iz}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{2}{z}_{\mathrm{1}} −{iz}_{\mathrm{1}} −\mathrm{2}+{z}_{\mathrm{1}} −\mathrm{2}{i}+{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:{iz}_{\mathrm{1}} ^{\mathrm{2}} +\left(\mathrm{3}−{i}\right){z}_{\mathrm{1}} −\left(\mathrm{2}−{r}^{\mathrm{2}} +\mathrm{2}{i}\right)=\mathrm{0} \\ $$$${as}\:{z}_{\mathrm{1}} \:{is}\:{a}\:{double}\:{root}\:\left({line}\:{is}\right. \\ $$$$\left.{tangent}\:{to}\:{circle}\right)\:{we}\:{must}\:{have} \\ $$$${D}=\mathrm{0}\:,\:{implies} \\ $$$$\:\:\:\left(\mathrm{3}−{i}\right)^{\mathrm{2}} +\mathrm{4}{i}\left(\mathrm{2}−{r}^{\mathrm{2}} +\mathrm{2}{i}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\mathrm{8}−\mathrm{6}{i}+\mathrm{8}{i}−\mathrm{4}{ir}^{\mathrm{2}} −\mathrm{8}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\mathrm{4}{r}^{\mathrm{2}} \:=\mathrm{2}\:\:\:{or}\:\:\:\:\boldsymbol{{r}}=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:. \\ $$$${Hence}\:{equation}\:{of}\:{circle}\:{is} \\ $$$$\:\:\:\:\:\mid\boldsymbol{{z}}−\mathrm{1}−\boldsymbol{{i}}\mid=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\ $$$${or}\:\:\left({z}−\mathrm{1}−{i}\right)\left(\bar {{z}}−\mathrm{1}+{i}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${z}\bar {{z}}+\left({i}−\mathrm{1}\right){z}−\left(\mathrm{1}+{i}\right)\bar {{z}}+\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\boldsymbol{{z}}\bar {\boldsymbol{{z}}}+\left(\boldsymbol{{i}}−\mathrm{1}\right)\boldsymbol{{z}}−\left(\mathrm{1}+\boldsymbol{{i}}\right)\bar {\boldsymbol{{z}}}+\mathrm{3}/\mathrm{2}=\mathrm{0}\:. \\ $$

Commented by Tinkutara last updated on 30/Aug/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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