The equation of the circle having its centre | vedclass

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(A) The family of circles passing through the intersection of two circles ${S_1} = 0$ and ${S_2} = 0$ is given by ${S_1} + \lambda {S_2} = 0$ (where $

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${x^2} + {y^2} - 6x + 7 = 0$

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(A) The family of circles passing through the intersection of two circles ${S_1} = 0$ and ${S_2} = 0$ is given by ${S_1} + \lambda {S_2} = 0$ (where $\lambda 
eq -1$).Given ${S_1}: {x^2} + {y^2} - 2x - 4y + 1 = 0$ and ${S_2}: {x^2} + {y^2} - 4x - 2y + 4 = 0$.The equation is: $({x^2} + {y^2} - 2x - 4y + 1) + \lambda ({x^2} + {y^2} - 4x - 2y + 4) = 0$.$(1 + \lambda ){x^2} + (1 + \lambda ){y^2} - (2 + 4\lambda )x - (4 + 2\lambda )y + (1 + 4\lambda ) = 0$.Dividing by $(1 + \lambda )$: ${x^2} + {y^2} - 2\frac{(1 + 2\lambda )}{1 + \lambda }x - 2\frac{(2 + \lambda )}{1 + \lambda }y + \frac{1 + 4\lambda }{1 + \lambda } = 0$.The centre of this circle is $\left( \frac{1 + 2\lambda }{1 + \lambda }, \frac{2 + \lambda }{1 + \lambda } ight)$.Since the centre lies on $x + 2y - 3 = 0$,we substitute the coordinates:$\frac{1 + 2\lambda }{1 + \lambda } + 2\left( \frac{2 + \lambda }{1 + \lambda } ight) - 3 = 0$.$1 + 2\lambda + 4 + 2\lambda - 3(1 + \lambda ) = 0$.$5 + 4\lambda - 3 - 3\lambda = 0 \implies \lambda + 2 = 0 \implies \lambda = -2$.Substituting $\lambda = -2$ into the family equation:$({x^2} + {y^2} - 2x - 4y + 1) - 2({x^2} + {y^2} - 4x - 2y + 4) = 0$.$-{x^2} - {y^2} + 6x - 7 = 0 \implies {x^2} + {y^2} - 6x + 7 = 0$.

The equation of the circle having its centre on the line $x + 2y - 3 = 0$ and passing through the points of intersection of the circles ${x^2} + {y^2} - 2x - 4y + 1 = 0$ and ${x^2} + {y^2} - 4x - 2y + 4 = 0$ is:

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