JEE Main & Advanced
Sample Paper
JEE Main - Mock Test - 32
question_answer
Tangents AB and AC are drawn from the point A(0, 1) to the circle \[{{x}^{2}}+{{y}^{2}}-2x+4y+1=0,\] then equation of circle through A, B and C is -
A)\[{{x}^{2}}+{{y}^{2}}+x+y-2=0\]
B)\[{{x}^{2}}+{{y}^{2}}-x+y-2=0\]
C)\[{{x}^{2}}+{{y}^{2}}+x-y-2=0\]
D)None of these
Correct Answer:
B
Solution :
[b] Equation of chord of contact (BC) T=0 \[x(0)+y(1)-2\left( \frac{x+0}{2} \right)+4\left( \frac{y+1}{2} \right)+1=0\] \[\Rightarrow -x+3y+3=0\] Let equation of circle passes through A, B, C is \[S+\lambda L=0\] \[\Rightarrow \left( {{x}^{2}}+{{y}^{2}}-2x+4y+1 \right)+\lambda \left( -x+3y+3 \right)=0\] It passes through A (0, 1) \[\therefore \left( 0+1-0+4+1 \right)+\lambda \left( 0+3+3 \right)=0\] \[\lambda =-1\] so equation of the required circle is \[{{x}^{2}}+{{y}^{2}}-x+y-2=0\]