定理–只有一个圆穿过三个给定的非共线点|第9类数学

定理：只有一个圆穿过三个给定的非共线点。

必填图：

给定：三个非共线点P，Q和R。

证明：只有一个圆通过P，Q和R。

建设：加入PQ和QR。分别画出PQ和RQ的垂直等分线AL和BM。由于P，Q，R不是共线的。因此，垂直平分线AL和BM不平行。让AL和BM在O相交。加入OP，OQ和OR。

证明

Since O lies on the perpendicular bisector of PQ. Therefore,

OP = OQ 

Again, O lies on the perpendicular bisector of QR. Therefore,  

OQ = OR

Thus, OP = OQ = OR = r (say).

Taking O as the centre draw a circle of radius s. Clearly, C (0, s) passes through P, Q, and R. This proves that there is a circle passing through the points P, Q, and R. We shall now prove that this is the only circle passing through P. Q and R. If possible, let there be another circle with centre O’ and radius r, passing through the points P, Q and R. Then, O’ will lie on the perpendicular bisectors AL of PQ and BM of QR. Since two lines cannot intersect at more than one point, so O’ must coincide with O. Since OP = r, O’P = s and O and O’ coincide, which means that,

r = s

Therefore,

C(O, r) ≅ (O’, s)

Hence, there is one and only one circle passing through three non-collinear points P, Q and R.

例子

示例1：写下逐步的构造过程以找出圆心?

解决方案：

Let the circle be C1. We need to find its centre.

Step 1: Take points P, Q, R on the circle 

Step 2: Join PR and RQ.

We know that the perpendicular bisector of a chord passes through the centre. So, we construct perpendicular bisectors of PR and RQ.

Step 3: Take a compass. With point P as the pointy end and R as the pencil end of the compass, mark an arc above and below PR. Do the same with R as pointy end P as pencil end of the compass.

Step 4: Join points intersected by the arcs. The line formed is the perpendicular bisector of PR.

Step 5: Take a compass, with point R as the pointy end and Q as pencil end of the compass mark an arc above and below RQ. Do the same with Q as the pointy end and R as the pencil end of the compass.

Step 6: Join the points intersected by the arcs. The line formed is the perpendicular bisector of RQ.

Step 7: The point where two perpendicular bisectors intersect is the centre of the circle. Mark it as point O. Thus, O is the centre of the given circle.

示例2：绘制不同的圆对。每对有多少个共同点?共同点的最大数量是多少?

解决方案：

(i)没有共同点：

(ii)共同点：

(iii)两点共同点：

正如我们可以从上面进行分析的那样，两个圆可以在两个点处彼此最大割。