From the points marking the ends of the diameter we can find its length and hence the radius of the circle and its centre. The difference between the y values is 2-(-4)=6, and between the x values is 5-(-3)=8. These form two sides of a right-angled triangle with the diameter as hypotenuse=sqrt(36+64)=10. So the radius is 5 and the centre is at ((5-3)/2, (2-4)/2)=(1,-1), therefore using the standard equation for a circle and substituting we get (x-1)^2+(y+1)^2=25 (standard eqn of a circle is (x-h)^2+(y-k)^2=a^2, where (h,k) is the centre and a the radius). The point (1,4) lies on the circle.
Another more long-winded solution is to find h, k and a by solving three equations generated by substituting the three given points. If this method is used it doesn't require the information that two points mark the ends of the diameter. The question, in fact, has too much information.