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Question

The radius of the circle x2+y22x+3y+k=0 is 212. Find the value of k. Find also the equation of the diameter of the circle which passes through the point (5,212).

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Solution

x2+y22x+3y+k=0
Radius =2.5
Comparing this equation with x2+y2+2gx+2fy+c=0, we get
g=1,f=1.5,c=k
Centre of circle =(g,f)=(1,1.5)
Radius of circle =g2+f2c
=> 2.5=(1)2+(1.5)2k
=> k=3

Slope of line passing through center (1,1.5) and point (5,212) is
m=2.5(1.5)51 => m=1
Now equation of line passing through (1,1.5) and having slope 1 is
(yy1)=m(xx1)
=> (y+1.5)=1(x1)
=> xy2.5=0

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