1. **State the problem:** We are given points $A(2,5)$, $B(-6,-3)$, and $C(a,1)$. We need to find the possible values of $a$ such that the length of $AB$ is twice the length of $AC$. 2. **Recall the distance formula:** The distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ 3. **Calculate length of $AB$:** $$AB = \sqrt{(-6 - 2)^2 + (-3 - 5)^2} = \sqrt{(-8)^2 + (-8)^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2}$$ 4. **Calculate length of $AC$:** $$AC = \sqrt{(a - 2)^2 + (1 - 5)^2} = \sqrt{(a - 2)^2 + (-4)^2} = \sqrt{(a - 2)^2 + 16}$$ 5. **Set up the equation given the condition:** Length of $AB$ is twice length of $AC$, so $$AB = 2 \times AC$$ Substitute the values: $$8\sqrt{2} = 2 \times \sqrt{(a - 2)^2 + 16}$$ 6. **Solve for $a$:** Divide both sides by 2: $$4\sqrt{2} = \sqrt{(a - 2)^2 + 16}$$ Square both sides: $$\left(4\sqrt{2}\right)^2 = (a - 2)^2 + 16$$ $$16 \times 2 = (a - 2)^2 + 16$$ $$32 = (a - 2)^2 + 16$$ Subtract 16: $$32 - 16 = (a - 2)^2$$ $$16 = (a - 2)^2$$ Take square root: $$a - 2 = \pm 4$$ 7. **Find values of $a$:** If $a - 2 = 4$, then $a = 6$ If $a - 2 = -4$, then $a = -2$ **Final answer:** The possible values of $a$ are $6$ and $-2$.