If the circle x^2+y^2-6x+2y=28 cuts off a cho | vedclass

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(B) Given circle: $x^2+y^2-6x+2y-28=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3, f=1, c=-28$. Centre $O = (-g, -f) = (3, -1)$. Radius $r = \

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$6$

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(B) Given circle: $x^2+y^2-6x+2y-28=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3, f=1, c=-28$. Centre $O = (-g, -f) = (3, -1)$. Radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-3)^2+(1)^2-(-28)} = \sqrt{9+1+28} = \sqrt{38}$ units. Let $OD$ be the perpendicular distance from the centre $O(3, -1)$ to the line $2x-5y+18=0$. $OD = \left|\frac{2(3)-5(-1)+18}{\sqrt{2^2+(-5)^2}}ight| = \left|\frac{6+5+18}{\sqrt{4+25}}ight| = \frac{29}{\sqrt{29}} = \sqrt{29}$ units. In $	riangle OAD$,by Pythagoras theorem: $AD^2 = r^2 - OD^2 = (\sqrt{38})^2 - (\sqrt{29})^2 = 38 - 29 = 9$. $AD = 3$ units. Since the perpendicular from the centre to a chord bisects the chord,the length of the chord $\lambda = AB = 2AD = 2 	imes 3 = 6$ units.

Column $I$	Column $II$
$(A)$ Two intersecting circles	$(p)$ have a common tangent
$(B)$ Two mutually external circles	$(q)$ have a common normal
$(C)$ Two circles,one strictly inside the other	$(r)$ do not have a common tangent
$(D)$ Two branches of a hyperbola	$(s)$ do not have a common normal

If the circle $x^2+y^2-6x+2y=28$ cuts off a chord of length $\lambda$ units on the line $2x-5y+18=0$,then the value of $\lambda$ is

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