The radius of the circle x2+y2−2x+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+y2−2x+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+f2−c
=> 2.5=√(−1)2+(1.5)2−k
=> k=−3
Slope of line passing through center (1,−1.5) and point (5,212) is
m=2.5−(−1.5)5−1 => m=1
Now equation of line passing through (1,−1.5) and having slope 1 is