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Equation of tangent to the circle $x^{2}+y^{2}-4x-16=0$ at point (0,4) is
Option: 1
2x-y+16=0

Option: 2
2x+4y+16=0

Option: 3
2x-4y-16=0

Option: 4
2x-4y+16=0

Answers (1)

Equation of tangent to the circle $x^{2}+y^{2}-4x-16=0$ at point (0,4) is

4y-2(x+0)-16=0

2x-4y+16=0

Posted by

Ramraj Saini

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