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The tangent to the circle x^2+y^2-4x+2y+k=0 at (1,1) is x-2y+1=0 then k=

The tangent to the circle x^2+y^2-4x+2y+k=0 at (1,1) is x-2y+1=0 then k=

Grade:11

2 Answers

Arun
25750 Points
4 years ago
 
Dear student
 
tangent at (1,1) is
 
 x* 1 + y* 1 – 2( x +1) + (y + 1) + k = 0
 
 – x + 2y – 1 + k = 0
 
x – 2y + 1 – k = 0
on comparing
 
1 – k = 1
hence k = 0
Vikas TU
14149 Points
4 years ago
Dear student 
you can simply solve this question by this method :
Dear student 
(1,1) Lies on circle and tangent both. 
So, put point (1,1) in equation o circle 
1^2 + 1^2 -4(1) +2(1)+k =0 
4-4 + k =0 
k=0 
Hope this helps 

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