The equation of the circle whose centre lies | vedclass

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(C) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$. Since its centre $(-g, -f)$ lies on the line $x-4y=1$,we have $-g-4(-f)=1$,which simplifi

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$x^2+y^2+6x+2y-90=0$

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(C) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$. Since its centre $(-g, -f)$ lies on the line $x-4y=1$,we have $-g-4(-f)=1$,which simplifies to $-g+4f=1$ $\dots(i)$. Since the circle passes through $(3,7)$,we have $3^2+7^2+2g(3)+2f(7)+c=0$,which gives $6g+14f+c=-58$ $\dots(ii)$. Since the circle passes through $(5,5)$,we have $5^2+5^2+2g(5)+2f(5)+c=0$,which gives $10g+10f+c=-50$ $\dots(iii)$. Subtracting equation $(ii)$ from $(iii)$,we get $(10g-6g)+(10f-14f) = -50 - (-58)$,so $4g-4f=8$,or $g-f=2$ $\dots(iv)$. Solving $(i)$ and $(iv)$: adding them gives $3f=3$,so $f=1$. Substituting $f=1$ in $(iv)$,we get $g=3$. Substituting $g=3$ and $f=1$ in $(iii)$,we get $10(3)+10(1)+c=-50$,so $30+10+c=-50$,which gives $c=-90$. Thus,the equation of the circle is $x^2+y^2+6x+2y-90=0$.

The equation of the circle whose centre lies on the line $x-4y=1$ and which passes through the points $(3,7)$ and $(5,5)$ is

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