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 
(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