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

The general equation of a circle in the plane is $$x^{2}+y^{2}+2gx+2fy+c=0$$, where

center = $$(-g,\,-f)$$ and radius $$r=\sqrt{\,g^{2}+f^{2}-c\,}$$.

Because three non-collinear points determine a unique circle, first obtain the circle passing through the points $$(4,6)$$, $$(-1,5)$$ and $$(0,0)$$.

Substituting $$(4,6)$$ gives
$$16+36+8g+12f+c=0 \quad\Rightarrow\quad 52+8g+12f+c=0 \; -(1)$$

Substituting $$(-1,5)$$ gives
$$1+25-2g+10f+c=0 \quad\Rightarrow\quad 26-2g+10f+c=0 \; -(2)$$

Substituting $$(0,0)$$ gives
$$c=0 \; -(3)$$

With $$c=0$$, equations $$(1)$$ and $$(2)$$ reduce to

$$52+8g+12f=0 \; -(4)$$
$$26-2g+10f=0 \; -(5)$$

Solve $$(4)$$ and $$(5)$$ simultaneously:

Multiply $$(5)$$ by $$4$$: $$104-8g+40f=0$$.
Add to $$(4)$$: $$ (52+104) + (12f+40f) = 0 \;\Rightarrow\; 156+52f=0$$,
so $$f=-3$$.

Insert $$f=-3$$ into $$(4)$$:
$$52+8g+12(-3)=0 \;\Rightarrow\; 52+8g-36=0 \;\Rightarrow\; 8g=-16 \;\Rightarrow\; g=-2$$.

Hence the circle through the three known points is
$$x^{2}+y^{2}-4x-6y=0$$.

Its center is $$(-g,\,-f)=(2,3)$$ and the radius is

$$r=\sqrt{g^{2}+f^{2}-c}=\sqrt{(-2)^{2}+(-3)^{2}-0}=\sqrt{4+9}= \sqrt{13}$$,
so $$r^{2}=13$$.

Now impose the condition that the fourth point $$(k,\,3k)$$ also satisfies the circle equation:

$$k^{2}+(3k)^{2}-4k-6(3k)=0$$
$$\Longrightarrow\; k^{2}+9k^{2}-4k-18k=0$$
$$\Longrightarrow\; 10k^{2}-22k=0$$
$$\Longrightarrow\; 2k\,(5k-11)=0$$.

Because the four points are distinct, $$k \neq 0$$; therefore

$$5k-11=0 \;\Rightarrow\; k=\frac{11}{5}$$.

Finally, evaluate $$10k+r^{2}$$:

$$10k+r^{2}=10\left(\frac{11}{5}\right)+13=22+13=35$$.

Hence $$10k+r^{2}=35$$, which corresponds to Option D.