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Question 1106619: Find the equation of a circle passing through two points (-4,1) and (4,5) whose centre lies on the straight line 2x-y=1
Answer by dkppathak(439)   (Show Source):
You can put this solution on YOUR website! Find the equation of a circle passing through two points (-4,1) and (4,5) whose centre lies on the straight line 2x-y=1
(X-h)^2 +(y-k)^2=r^2
the circle passing through -4,1
(-4-h)^2 +(1-k)^2=r^2 ---(1)
the circle passing through 4,5
(4-h)^2 +(5-k)^2=r^2 ---(2) 1 and 2 both are equal
(-4-h)^2 +(1-k)^2 =(4-h)^2 +(5-k)^2
16+h^2+8h+1+k^2-2k=16+h^2-8h+25+k^2-10k
16h+8k-24=0
2h+k=3 (3)
center lies on line 2x-y=1 center h,k lies and satisfy the line
2h-k=1 (4) by solving 3, 4 we will get h=1 and k=1
by putting the value of h and k in equation 1 we will get radius =5
equation of circle will bw
(x-1)^2 +(y-1)^2=25
or
x^2+y^2-2x-2y-23=0
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