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Question Number 174315 by mnjuly1970 last updated on 29/Jul/22

	Answered by aleks041103 last updated on 29/Jul/22

	$i f t h e i n t e r s e c t i o n p o i n t o n A B i s O,$ ${l e t}^{'} s s e t i t s c o o r d i n a t e s t o (0, 0)$ $t h e n O B i s t h e x a x i s .$ $t h e n t h e q u a t i o n o f t h e s l a n t e d l i n e i s :$ $y = x$ $a n d t h e e q n . o f t h e c i r c l e i s :$ ${(x - 2)}^{2} + y^{2} = 100$ $t h e i n t e r s e c t i o n i s$ ${(x - 2)}^{2} + x^{2} = 100$ $\Rightarrow 2 x^{2} - 4 x + 4 = 100$ $x^{2} - 2 x + 2 = 50$ $x^{2} - 2 x - 48 = 0$ $x^{2} - 8 x + 6 x - 48 = 0$ $x (x - 8) + 6 (x - 8) = 0$ $\Rightarrow (x + 6) (x - 8) = 0$ $\Rightarrow x_{i n t e r s e c} = + 8, - 6$ $s i n c e t h e l i n e i s s l a n t e d a t 45 °, t h e n :$ $x = 8 \sqrt{2}, y = 6 \sqrt{2}$ $\Rightarrow x + y = 14 \sqrt{2}$
	Commented by Tawa11 last updated on 29/Jul/22

	$Great sir$
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