If the four distinct points (4,6), (-1,5), (0 | vedclass

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(D) Let the points be $A(4,6), B(-1,5), C(0,0)$ and $D(k, 3k)$.First,check if $	riangle ABC$ is a right-angled triangle. The slopes of $AC$ and $BC$

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$35$

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(D) Let the points be $A(4,6), B(-1,5), C(0,0)$ and $D(k, 3k)$.First,check if $	riangle ABC$ is a right-angled triangle. The slopes of $AC$ and $BC$ are $m_{AC} = \frac{6-0}{4-0} = \frac{3}{2}$ and $m_{BC} = \frac{5-0}{-1-0} = -5$. This does not give a right angle. However,checking $AB$ and $BC$: $m_{AB} = \frac{5-6}{-1-4} = \frac{-1}{-5} = \frac{1}{5}$ and $m_{BC} = -5$. Since $m_{AB} \cdot m_{BC} = -1$,$\angle ABC = 90^\circ$.Thus,$AC$ is the diameter of the circle passing through $A, B, C$. The equation of the circle with diameter $AC$ is $(x-4)(x-0) + (y-6)(y-0) = 0$,which simplifies to $x^2 + y^2 - 4x - 6y = 0$.The point $D(k, 3k)$ lies on this circle,so $k^2 + (3k)^2 - 4k - 6(3k) = 0$.$10k^2 - 22k = 0 \implies 2k(5k - 11) = 0$. Since the points are distinct,$k 
eq 0$,so $k = \frac{11}{5}$.The center of the circle is the midpoint of $AC$,which is $(\frac{4+0}{2}, \frac{6+0}{2}) = (2, 3)$.The radius squared is $r^2 = (2-0)^2 + (3-0)^2 = 4 + 9 = 13$.Finally,$10k + r^2 = 10(\frac{11}{5}) + 13 = 22 + 13 = 35$.

If the four distinct points $(4,6), (-1,5), (0,0)$ and $(k, 3k)$ lie on a circle of radius $r$,then $10k + r^2$ is equal to

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