The equation of the circle having its centre | vedclass

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Question

(a) Required circle will be ${S_1} + \lambda {S_2} = 0$, $\lambda 
e - 1$

i.e., ${x^2} + {y^2} - 2x - 4y + 1 + \lambda ({x^2} + {y^2} - 4x - 2y +

Vedclass · Accepted Answer

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

Vedclass · Answer

(a) Required circle will be ${S_1} + \lambda {S_2} = 0$, $\lambda 
e - 1$

i.e., ${x^2} + {y^2} - 2x - 4y + 1 + \lambda ({x^2} + {y^2} - 4x - 2y + 4) = 0$

==> ${x^2} + {y^2} - 2\frac{{(1 + 2\lambda )}}{{1 + \lambda }}x - 2\frac{{(2 + \lambda )}}{{1 + \lambda }}y + \frac{{1 + 4\lambda }}{{1 + \lambda }} = 0$

Its centre $\left( {\frac{{1 + 2\lambda }}{{1 + \lambda }},\;\frac{{2 + \lambda }}{{1 + \lambda }}} ight)$ lies on $x + 2y - 3 = 0$

 $\frac{{1 + 2\lambda }}{{1 + \lambda }} + 2\left( {\frac{{2 + \lambda }}{{1 + \lambda }}} ight) - 3 = 0$

==> $\lambda = - 2$.

 The circle is ${x^2} + {y^2} - 6x + 7 = 0$

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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