If the midpoint of the chord intercepted by t | vedclass

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(A) The equation of the circle is $x^2+y^2-8x+10y+5=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=-4$ and $f=5$. The center of the circle is

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(A) The equation of the circle is $x^2+y^2-8x+10y+5=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=-4$ and $f=5$. The center of the circle is $C(-g, -f) = (4, -5)$.Let the midpoint of the chord be $P(h, k)$. Since $P$ lies on the line $2x+y+2=0$,we have $2h+k+2=0$ ... $(i)$.The line segment $CP$ is perpendicular to the chord $2x+y+2=0$. The slope of the chord is $-2$. Therefore,the slope of $CP$ is $\frac{1}{2}$.The slope of $CP$ is given by $\frac{k-(-5)}{h-4} = \frac{k+5}{h-4}$.Equating the slopes: $\frac{k+5}{h-4} = \frac{1}{2}$ $\Rightarrow 2k+10 = h-4$ $\Rightarrow h-2k=14$ ... (ii).Solving equations $(i)$ and (ii):From $(i)$,$k = -2h-2$. Substituting into (ii): $h-2(-2h-2) = 14$ $\Rightarrow h+4h+4=14$ $\Rightarrow 5h=10$ $\Rightarrow h=2$.Then $k = -2(2)-2 = -6$.Thus,$k+4h = -6+4(2) = -6+8 = 2$.

If the midpoint of the chord intercepted by the circle $x^2+y^2-8x+10y+5=0$ on the line $2x+y+2=0$ is $(h, k)$,then $k+4h=$

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