A) \[3x-y-4=0\]
B) \[x+4y+3=0\]
C) \[x-3y-4=0\]
D) \[4x+y-3=0\]
Correct Answer: B
Solution :
Centre of circle is\[\left( \frac{4}{3},\frac{1}{3} \right)\] \[\Rightarrow \,\]equation of circle is \[{{\left( x-\frac{4}{3} \right)}^{2}}+{{\left( y-\frac{1}{3} \right)}^{2}}={{\left( 1-\frac{4}{3} \right)}^{2}}+{{\left( -1-\frac{1}{3} \right)}^{2}}\] \[\Rightarrow \,\,{{x}^{2}}-\frac{8}{3}x+\frac{16}{9}+{{y}^{2}}-\frac{2}{3}y+\frac{1}{9}=\frac{1}{9}+\frac{16}{9}\] \[\Rightarrow \,\,{{x}^{2}}+{{y}^{2}}-\frac{8}{3}\times -\frac{2}{3}y=0\] \[\Rightarrow \,\,3{{x}^{2}}+3{{y}^{2}}-8x-2y=0\] Equation of tangent at \[(1,-1)\] is \[3x-3y-4(x+1)-(y-1)=0\] \[\Rightarrow \,-x-=4y-3=0\] \[\Rightarrow \,x+4y+3=0\]
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