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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-

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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.
Findtheequationofcircleincomplexformwhichtouchesiz+z¯+1+i=0andforwhichthelines(1−i)z=(1+i)z¯and(1+i)z+(i−1)z¯−4i=0arenormals.
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 .
Centerofcircleliesonboththenormals:letcenterbez0.⇒(1+i)z0−(1−i)z¯0=4i(1−i)z0−(1+i)z¯0=0adding:2z0−2z¯0=4iorz0−z¯0=2i….(i)subtracting:2iz0+2iz¯0=4iorz0+z¯0=2….(ii)adding(i)and(ii):z0=1+iHenceequationofcirclecanbeassumedtobe:∣z−z0∣=randletgiventangenttocircletouchesitatz1.Then∣z1−z0∣=rwhere(z0=1+i)or(z1−z0)(z¯1−z¯0)=r2…(iii)fromequationoftangentiz1+z¯1+1+i=0⇒z¯1=−(iz1+1+i)substitutingin(iii)withz0=1+i(z1−1−i)(−iz1−1−i−1+i)=r2(z1−1−i)(iz1+1+i+1−i)=−r2⇒(z1−1−i)(iz1+2)=−r2oriz12+2z1−iz1−2+z1−2i+r2=0iz12+(3−i)z1−(2−r2+2i)=0asz1isadoubleroot(lineistangenttocircle)wemusthaveD=0,implies(3−i)2+4i(2−r2+2i)=0⇒8−6i+8i−4ir2−8=0⇒4r2=2orr=12.Henceequationofcircleis∣z−1−i∣=12or(z−1−i)(z¯−1+i)=12zz¯+(i−1)z−(1+i)z¯+2−12=0zz¯+(i−1)z−(1+i)z¯+3/2=0.
Commented by Tinkutara last updated on 30/Aug/17
Thank you very much Sir!
ThankyouverymuchSir!ThankyouverymuchSir!

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