Math Forum :: View topic – Circles……..

alphaboy wrote:

Still don’t know 2), 3) and 4)…………


From 緣’s solution, by equations (1) and (2), you just need to evaluate the 3 unknowns : h, k and r. It normally requies 3 equations. By substituting (0, -1) into (2), you get . When , . Its discriminant = which is the 2nd equation. You can get the 3rd equation in a similar manner when y=1.

An alternative method can be made by taking merit from geometric situation. If the circle locates at top right hand corner, then h=k=r+1. If it locates at bottom right hand corner, then h=r+1, k=r-1 and so on. Therefore, and . Sub. them into (1), . Here we only have 1 unknown r. But substituting (0, -1) into it, r can be easily solved.


As 緣 said, the centre is C(h, 0). Note that the distance between C(h, 0) and (1, 1) is equal to that between C(h, 0) and the line 2x-y-1=0, both are actually the length of radius. By using the formula of distance between a point and a line, we get . Hence, h can be solved. The radius will be \sqrt{(h-1)^2 + (0-1)^2}. Once you get the centre and radius, you get the equation of the circle by equation (1) given by 緣.


Let the centre be C(h, k) and radius be r. Since the circle touches the x-axis, it implies . Hence, the equation of the circle becomes . You have 2 unknowns h and r. By substituting (-2,-1) ,(4,-1) into it, you will get two equations, hence h and r can be solved.

Work out yourself. Good luck!