Circle And System Of Circles Question 264
Question: Tangents AB and AC are drawn from the point $ A(0,,1) $ to the circle $ x^{2}+y^{2}-2x+4y+1=0 $ . Equation of the circle through A, B and C is
Options:
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
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Answer:
Correct Answer: B
Solution:
Equation of BC (chord of contact) is $ 0.x+1.y-(x+0)+2(y+1)+1=0 $ or $ -x+3y+3=0 $
Equation of circle through B and C i.e., intersection of the given circle and chord of contact is $ (x^{2}+y^{2}-2x+4y+1)+\lambda (-x+3y+3)=0 $ .
It passes through $ A(0,\ 1) $ , so the equation of the required circle is $ x^{2}+y^{2}-x+y-2=0 $ .
Aliter: Centre of the required circle is mid-point of $ A(0,\ 1) $ and centre of the given circle i.e., $ (1,\ -2) $ .
Therefore, centre $ ( \frac{1}{2},\ -\frac{1}{2} ) $ and radius $ \sqrt{\frac{5}{2}} $ .
Hence the circle is $ x^{2}+y^{2}-x+y-2=0 $ .
BETA
