Conswing line $x + y = 5$ and the circle $x^{2} + y^{2} - 2 x - 4 y + 3 = 0$ then

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Solution

$\begin{matrix} G i v e n l i n e : x + y - 5 = 0 c i r c l e e q u : x^{2} + y^{2} - 2 x - 4 y + 3 = 0 w e t a k e, 2 g x = - 2 x g = - 1 2 f y = - 4 y f = - 2 c e n t r e (- g - f) = (1, 2) r a d i u s : \sqrt{g^{2} + f^{2} - c} = \sqrt{1 + 4 - 30} = \sqrt{2} N o w, p e r p e n d i c u l a r f r o m c e n t r e t o x + y - 5 = 0 e q u a l t o r a d i u s o f c i r c l e . P = \frac{| 1 + 2 - 5 |}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} h e n c e, t h e l i n e t o u c h e s i n t h e c i r c l e . l e t t h e p o i n t (h, k) a n d c e n t r e b e C, s o, P C ⊥ t o x + y - 5 = 0 ∣ ∣ P C = \frac{k - 2}{h - 1} \Rightarrow y = - x + 5 = - 1 \Rightarrow \frac{k - 2}{h - 1} \times - 1 = - 1 \Rightarrow k - 2 = h - 1 \Rightarrow k - h = 1 - - - - - - - - - - (I) a l s o h, k l i e o n t h e l i n e, s o : h + k = 5 - - - - - - - - - - (2) s o l v i n g e q u (1) & (2) 2 k = 6 k = 3, h = 2 H e n c e, p o i n t i s P (2, 3) \end{matrix}$

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