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  • What is the maximum distance of the point (3,4) from the circle is <img alt="x^{2}+y^{2}-4x-2y-4=0" src="https://learn.careers360.com/latex-image/?x%5E%7B2%7D&plus;y%5E%7B2%7D-4x-2y-4%3D0"

What is the maximum distance of the point (3,4) from the circle is x^{2}+y^{2}-4x-2y-4=0

Option: 1

9


Option: 2

11


Option: 3

3-2\sqrt2


Option: 4

3+2\sqrt2


Answers (1)

best_answer

First, check if the point (4,3) lies inside or outside of the circle

\left(4\right)^2+\left(3\right)^2-4\left(4\right)-2\left(3\right)-4=-1<0

P(4,3) lies inside the circle

center of the circle is C(-g,-f) = (2,1)

CP = \sqrt{(2-4)^2+(1-3)^2}=\sqrt{(-2)^2+(-2)^2}=2\sqrt2

radius of the circle is \sqrt{g^2+f^2-c}=\sqrt{2^2+1^2-(-4)}=3

Maximum diistance = CP+r = 3+2\sqrt2

Posted by

Rishabh

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