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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=

```
one year ago

Arun
25141 Points
```							 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 = 0on comparing 1 – k = 1hence k = 0
```
one year ago
Vikas TU
13786 Points
```							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
```
one year ago
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### Course Features

• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions