Find the equation of the circle which has its centre on the line y=2 and which passes through the points (2, 0) and (4, 0).

Dear student
Let the equation of the circle be x-h2+y-k2=r2Since the circle passes through 2,0 and (4,0), we have2-h2+0-k2=r2    ...(1)and 4-h2+0-k2=r2  ...(2)Also since the centre lies on the line y=2, we havek=2    ...(3)Putting the value of k in (1), we get2-h2+-22=r24+h2-4h+4=r2h2-4h+8=r2     ....(4)Putting the value of k in(2), we get4-h2+-22=r216+h2-8h+4=r2h2-8h+20=r2    ...(5)Equating (4) and (5), we geth2-4h+8=h2-8h+204h=12h=3Putting the value of h in (4), we get32-43+8=r2r2=5Hence the required equation of the circle is x-32+y-22=5x2+9-6x+y2+4-4y=5x2+y2-6x-4y+8=0
Regards

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