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Q. The equation of circle touches the line y=x at origin and passes through the point (2,1) is x²+y²+px+gy=0, then p+q is equal to-
1) 0
2) -10
3) 10
4) 5

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Solution

$C o n s i d e r a p o i n t c i r c l e c e n t e r e d a t o r i g i o n C : {(x - 0)}^{2} + {(y - 0)}^{2} = 0 x^{2} + y^{2} = 0 C o n s i d e r a l i n e L : y - x = 0 U \sin g f a m i l y o f c i r c l e s t h e c i r c l e p a s \sin g t h r o u g h i n t e r s e c t i o n o f C a n d L i s g i v e n b y x^{2} + y^{2} + λ (y - x) = 0 N o t e : T h i s c i r c l e p a s s e s t h r o u g h o n l y o n e p o i n t l y i n g o n L, i . e . (0, 0) . T h e r e f o r e i t t o u c h e s L . R e a s o n : F o r o t h e r p o i n t s l y i n g o n L, λ (y - x) w i l l b e z e r o b u t x^{2} + y^{2} w i l l y i e l d n o n - z e r o q u a n t i t y . I t i s g i v e n t h e c i r c l e p a s s e s t h r o u g h (2, 1) 2^{2} + 1^{2} + λ (1 - 2) = 0 5 - λ = 0 λ = 5 E q u a t i o n o f r e q u i r e d c i r c l e i s x^{2} + y^{2} + 5 (y - x) = 0 x^{2} + y^{2} + 5 y - 5 x = 0 A c o m p a r i n g w i t h x^{2} + y^{2} + p x + q y = 0 p = - 5 q = 5 p + q = 0$

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