Question 483350: The point (1,-1) lies on the circle whose equation is x2 + 4x +y2 - 6y = k. Find the value of k, the center and the radius of the circle.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! The point (1,-1) lies on the circle whose equation is x2 + 4x +y2 - 6y = k. Find the value of k, the center and the radius of the circle.
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x^2 + 4x +y^2 - 6y = k
plug in (x,y) coordinates of given point (1,-1)
k=1^2+4*1+(-1)^2-6*(-1)
=1+4+1+6=12
=x^2+4x+y^2-6y =12
completing the square
(x^2+4x+4)+(y^2-6y+9)=12+4+9=25
(x+2)^2+(y-3)^2=25
This is an equation of a circle of standard form: (x-h)^2+(y-k)^2=r^2, (h,k) being the (x,y) coordinates of the center, and r=radius.
..
For given equation:
k=12
center: (-2,3)
radius: √25=5
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