| Statement 1: Consider two curves \[{{C}_{l}}:z\overline{z}+i\overline{z}+b=0\] and \[(b\in R,z=x+iy\] and \[i=\sqrt{-1})\]. where\[(b\in R,\,z=x+iy\] and \[i=\sqrt{-1}\]). If \[{{C}_{1}}\] and \[{{C}_{2}}\] intersects orthogonally then\[b=-2\]. |
| Statement 2: If two curves intersects orthogonally then the angle between the tangents at all their points of intersection is\[\frac{\pi }{2}\]. |
A) Statement-1 is true, Statement-2 is true and Statement-2 is correct explanation for Statement-1.
B) Statement-1 is true, Statement-2 is true and Statement-2 is NOT the correct explanation for Statement-1.
C) Statement-1 is true, Statement-2 is false.
D) Statement-1 is false, Statement-2 is true
Correct Answer: D
Solution :
\[{{C}_{1}}\,:z\bar{z}+iz\,-i\,\bar{z}+b=0\] Put \[z=x+iy\] \[{{x}^{2}}+{{y}^{2}}\,+i(x+yi)-i(x-iy)+b=0\] \[\Rightarrow \,{{x}^{2}}+{{y}^{2}}\,-2y+b=0\] ?(1) Also \[z\bar{z}+(1-i)\,z+(1+i)\,\bar{z}-4=0\] \[\Rightarrow \,\,{{x}^{2}}+{{y}^{2}}\,+(1-i)\,(x+iy)\,+(1+i)\,(x-iy)\,-4=0\] \[\Rightarrow \,\,{{x}^{2}}+{{y}^{2}}\,+2x+2y-4=0\] ?(2) \[\therefore \] Using condition of orthogonally, we get \[2(0\times 1+(-1)(1))\,=b-4\,\Rightarrow \,-2=b-4\Rightarrow \,b=2\] \[\therefore \,\,S-1\] is false, but S-2 is true.
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