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find the equation of a circle with center (-4,2) and tangent to the line 2x-y+2=0


Sagot :

Given:
center of circle: (-4,2)
[tex]y=2(\frac{4}{5})+2[/tex]

Solution:
Using the point slope equation and the fact that perpendicular lines are negative reciprocals of each other. 

y - 2 = (-1/2)(x+4)
2y - 4 = -x - 4
2y = -x
[tex]y = \frac{-x}{2}[/tex]

Since the equation above is the equation of the line perpendicular to y=2x+2, we can find the point of intersection

[tex]2x + 2=\frac{-x}{2}[/tex]
4x + 4 = -x 
4x + x = -4
[tex]x = \frac{-4}{5}[/tex]

Subtstituting x in the give equation you get,
[tex]y=2(\frac{-4}{5})+2[/tex]
[tex]x = \frac{2}{5}[/tex]

Using the distance formula you get the radius of the circle.
[tex]r = \sqrt{(x_{2}-x_{1})^2+(y_{2} - y_{1})^2 } [/tex]