Grade 11IIT JEE Entrance Exam
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=
2 Answers
Arun
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
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
(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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