Suppose that the circle x^2+y^2+2gx+2fy+c=0 h | vedclass

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(D) The equation of the given circle is $x^2+y^2+2gx+2fy+c=0$ ...$(i)$ with centre $(-g, -f)$.Since the centre lies on the line $2x+3y-7=0$,we have $2

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$9$

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(D) The equation of the given circle is $x^2+y^2+2gx+2fy+c=0$ ...$(i)$ with centre $(-g, -f)$.Since the centre lies on the line $2x+3y-7=0$,we have $2(-g)+3(-f)-7=0$,which implies $2g+3f+7=0$ ...(ii).Two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ cut orthogonally if $2g_1g_2+2f_1f_2=c_1+c_2$.For the first circle $x^2+y^2-4x-6y+11=0$,we have $2g(-2)+2f(-3)=c+11$,which simplifies to $4g+6f+c+11=0$ ...(iii).For the second circle $x^2+y^2-10x-4y+21=0$,we have $2g(-5)+2f(-2)=c+21$,which simplifies to $10g+4f+c+21=0$ ...(iv).Subtracting (iii) from (iv),we get $(10g-4g)+(4f-6f)+(21-11)=0$,which is $6g-2f+10=0$,or $3g-f+5=0$ ...$(v)$.From (ii),$2g+3f=-7$. From $(v)$,$f=3g+5$. Substituting into (ii): $2g+3(3g+5)=-7$ $\Rightarrow 11g+15=-7$ $\Rightarrow 11g=-22$ $\Rightarrow g=-2$.Then $f=3(-2)+5=-1$. Substituting $g=-2, f=-1$ into (iii): $4(-2)+6(-1)+c+11=0$ $\Rightarrow -8-6+c+11=0$ $\Rightarrow c-3=0$ $\Rightarrow c=3$.Finally,$5g-10f+3c = 5(-2)-10(-1)+3(3) = -10+10+9 = 9$.

Suppose that the circle $x^2+y^2+2gx+2fy+c=0$ has its centre on $2x+3y-7=0$ and cuts the circles $x^2+y^2-4x-6y+11=0$ and $x^2+y^2-10x-4y+21=0$ orthogonally. Then $5g-10f+3c=$

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