x²+y²-4x-3等于0
x²+y²-4y-3等于0
x-y-4=0 --(1)
x^+y^2-4x-3=0 ---(2)
x^+y^2-4y-3=0 ---(3)
(3)-(2)=>
4x-4y=0
x=y
and we can get 2 points, let say(x1,y1)
and (x2,y2)
centre of the required circle is on (1), let the centre be(a,4+a)
and the equation of the circle is
(x-a)^2+(y-(4+a))^2 = r^2 ---(3)
where r be the radius of the circle.
Sub (x1,y1), (x2,y2) into (3), we can solve for a and r, then we can get the equation of the required circle.
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