The equation of a circle,which passes through | vedclass

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(D) The given circle is concentric with $2x^2+2y^2-8x-12y-9=0$,which simplifies to $x^2+y^2-4x-6y-4.5=0$. Thus,the equation of the required circle is

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$x^2+y^2-4x-6y-87=0$

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(D) The given circle is concentric with $2x^2+2y^2-8x-12y-9=0$,which simplifies to $x^2+y^2-4x-6y-4.5=0$. Thus,the equation of the required circle is of the form $x^2+y^2-4x-6y+k=0$. This circle passes through the centre of $x^2+y^2+8x+10y-7=0$. The centre of $x^2+y^2+8x+10y-7=0$ is $(-4, -5)$. Substituting $(-4, -5)$ into the equation $x^2+y^2-4x-6y+k=0$: $(-4)^2+(-5)^2-4(-4)-6(-5)+k=0$ $16+25+16+30+k=0$ $87+k=0 \Rightarrow k=-87$. Therefore,the required equation is $x^2+y^2-4x-6y-87=0$.

The equation of a circle,which passes through the centre of the circle $x^2+y^2+8x+10y-7=0$ and is concentric with the circle $2x^2+2y^2-8x-12y-9=0$,is

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