If the length of the chord 2x+3y+k=0 of the c | vedclass

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(C) The given equation of the circle is $x^2+y^2-6x-8y+9=0$.Its centre is $C=(3,4)$ and radius $R=\sqrt{3^2+4^2-9}=\sqrt{9+16-9}=\sqrt{16}=4$.Let $AB$

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$-5$

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(C) The given equation of the circle is $x^2+y^2-6x-8y+9=0$.Its centre is $C=(3,4)$ and radius $R=\sqrt{3^2+4^2-9}=\sqrt{9+16-9}=\sqrt{16}=4$.Let $AB$ be the chord of length $2\sqrt{3}$. The perpendicular distance $CM$ from the centre $C(3,4)$ to the chord $2x+3y+k=0$ is given by $CM = \frac{|2(3)+3(4)+k|}{\sqrt{2^2+3^2}} = \frac{|6+12+k|}{\sqrt{4+9}} = \frac{|18+k|}{\sqrt{13}}$.In the right-angled triangle $	riangle ACM$,we have $AC^2 = CM^2 + AM^2$,where $AM = \frac{AB}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3}$.Substituting the values,$4^2 = \left(\frac{|18+k|}{\sqrt{13}}ight)^2 + (\sqrt{3})^2$.$16 = \frac{(18+k)^2}{13} + 3$.$13 = \frac{(18+k)^2}{13} \Rightarrow (18+k)^2 = 169$.Taking the square root,$18+k = \pm 13$.Case $1$: $18+k = 13 \Rightarrow k = -5$.Case $2$: $18+k = -13 \Rightarrow k = -31$.Thus,one of the values of $k$ is $-5$.

If the length of the chord $2x+3y+k=0$ of the circle $x^2+y^2-6x-8y+9=0$ is $2\sqrt{3}$,then one of the values of $k$ is

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