The locus of the centres of the circles passing through the intersection of the circles $x^{2} + y^{2} = 1$ and $x^{2} + y^{2} - 2 x + y = 0$ is

a line whose equation is

x + 2 y = 0

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a line whose equation is

2 x - y = 1

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a circle

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a pair of line

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Solution

The correct option is C a line whose equation is $x + 2 y = 0$
equation of of the circles passing through the intersection of the circles $x^{2} + y^{2} = 1$ and $x^{2} + y^{2} - 2 x + y = 0$ is $x^{2} + y^{2} - 1 + k (x^{2} + y^{2} - 2 x + y) = 0$
$\Rightarrow x^{2} + y^{2} - (\frac{2 k}{k + 1}) x + (\frac{k}{k + 1}) y - 1 = 0$
center of circle is $(x, y) = (\frac{k}{k + 1}, - \frac{k}{2 (k + 1)})$
Therefore, locus is $x + 2 y = 0$
Ans: A

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The locus of the centres of the circles passing through the intersection of the circles x2+y2=1 and x2+y2−2x+y=0 is

The correct option is C a line whose equation is x+2y=0equation of of the circles passing through the intersection of the circles x2+y2=1 and x2+y2−2x+y=0 is x2+y2−1+k(x2+y2−2x+y)=0⇒x2+y2−(2kk+1)x+(kk+1)y−1=0center of circle is (x,y)=(kk+1,−k2(k+1))Therefore, locus is x+2y=0Ans: A

The locus of the centres of the circles passing through the intersection of the circles $x^{2} + y^{2} = 1$ and $x^{2} + y^{2} - 2 x + y = 0$ is