Jambmaths question:
A predator moves in a circle of radius $\sqrt{2}$centre (0,0), while a prey moves along y = x. If $0\le x\le 2$, at which point will they meet
Option A:
(0,0) and (1,1)
Option B:
(1,1) and (1,2)
Option C:
(1,1) only
Option D:
$(\sqrt{2},\sqrt{2})$only
Jamb Maths Solution:
$\begin{align} & {{x}^{2}}+{{y}^{2}}={{(\sqrt{2})}^{2}}\text{ }\!\!\{\!\!\text{ equation of circle }\!\!\}\!\!\text{ } \\ & {{x}^{2}}+{{y}^{2}}=2-----(i) \\ & y=x-------(ii) \\ & \text{Substitute }y=x\text{ in }(i) \\ & {{x}^{2}}+{{x}^{2}}=2 \\ & 2{{x}^{2}}=2 \\ & {{x}^{2}}=1 \\ & x=\sqrt{1}=\pm 1\text{ (ignore }x=-1\text{, it is outside the range of }x\} \\ & x=1 \\ & \text{since }y=x,y=1 \\ & (x,y)=(1,1) \\\end{align}$The predator and the prey will meet only at point (1,1). Option C is the correct option.
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