题面
解析
大概是圆反演的模板吧。
以点$P(x3, y3)$为反演中心，任意长为反演半径，将两个已知圆反演，设反演后的圆为$A'$, $B'$，所求圆反演后为一条直线，根据题目中的要求，该直线为两圆的外公切线。因此我们只需要求出两圆的外公切线即可。
然后会发现WA了，因为题目中还有一个要求，所求圆要外切于两圆，即反演变换后反演中心$P$和$A'$的圆心要在同侧。
还有一个我一开始做错了的地方，原来的圆心$O$反演后就不是新的圆心了！！！可以连接$PO$，求其与圆的两个交点，两个交点的反演点中点才是新的圆心。
代码：
#include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<cmath> using namespace std; const double eps = 1e-11, Pi = 3.14159265358979323; int T, cnt, num; double R; struct node{ double fir, sec; node(){} node(double x, double y) { fir = x; sec = y; } }p, c[5], d[5]; node operator + (node x, node y) { return node(x.fir + y.fir, x.sec + y.sec); } node operator - (node x, node y) { return node(x.fir - y.fir, x.sec - y.sec); } node operator * (node x, double y) { return node(x.fir * y, x.sec * y); } node operator / (node x, double y) { return node(x.fir / y, x.sec / y); } //Rotate vector counterclockwise by alpha node rotate(node x, double y)//rad { return node(x.fir * cos(y) - x.sec * sin(y), x.fir * sin(y) + x.sec * cos(y)); } //Cross product double crp(node x, node y) { return x.fir * y.sec - x.sec * y.fir; } //Dot product double dtp(node x, node y) { return x.fir * y.fir + x.sec * y.sec; } //Distance double dist(node x) { return sqrt(dtp(x, x)); } //Find the intersection of two straight lines //x, y are the vector starting points //u, v are the direction vectors node itpt(node x, node u, node y, node v) { node t = x - y; double rat = crp(v, t) / crp(u, v); return x + u * rat; } struct circ{ node o; double r; }a, b, aa, bb, ans[5]; //Find the common tangent of two circles void tagt(circ x, circ y) { if(x.r < y.r) swap(x, y); double len = dist(x.o - y.o); //External common tangent double alp = acos((x.r - y.r) / len); node tmp = y.o - x.o, t = rotate(tmp, alp); c[++cnt] = x.o + t / len * x.r; tmp = x.o - y.o; t = rotate(tmp, Pi + alp); d[cnt] = y.o + t / len * y.r; tmp = y.o - x.o; t = rotate(tmp, 2 * Pi - alp); c[++cnt] = x.o + t / len * x.r; tmp = x.o - y.o; t = rotate(tmp, Pi - alp); d[cnt] = y.o + t / len * y.r; //Internal common tangent /*alp = acos((x.r + y.r) / len); tmp = y.o - x.o; t = rotate(tmp, alp); c[++cnt] = x.o + t / len * x.r; tmp = x.o - y.o; t = rotate(tmp, alp); d[cnt] = y.o + t / len * y.r; tmp = y.o - x.o; t = rotate(tmp, 2 * Pi - alp); c[++cnt] = x.o + t / len * x.r; tmp = x.o - y.o; t = rotate(tmp, 2 * Pi - alp); d[cnt] = y.o + t / len * y.r;*/ } int check(double x) { return (x > eps) - (x < -eps); } void out(double x) { printf("%.8f ", fabs(x) < 0.000000005? 0: x); } int main() { scanf("%d", &T); R = 1.0; double len1, len2; node tmp, t; while(T --) { scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &a.o.fir, &a.o.sec, &a.r, &b.o.fir, &b.o.sec, &b.r, &p.fir, &p.sec); len1 = dist(a.o - p); tmp = p + (a.o - p) / len1 * (len1 - a.r); len2 = dist(tmp - p); tmp = p + (tmp - p) / len2 * (R / len2); t = p + (a.o - p) / len1 * (len1 + a.r); len2 = dist(t - p); t = p + (t - p) / len2 * (R / len2); aa.o = (tmp + t) / 2; aa.r = dist(aa.o - tmp); len1 = dist(b.o - p); tmp = p + (b.o - p) / len1 * (len1 - b.r); len2 = dist(tmp - p); tmp = p + (tmp - p) / len2 * (R / len2); t = p + (b.o - p) / len1 * (len1 + b.r); len2 = dist(t - p); t = p + (t - p) / len2 * (R / len2); bb.o = (tmp + t) / 2; bb.r = dist(bb.o - tmp); cnt = num = 0; tagt(aa, bb); for(int i = 1; i <= cnt; ++i) { if(fabs(crp(c[i] - p, d[i] - p)) < eps || check(crp(d[i] - c[i], p - c[i])) != check(crp(d[i] - c[i], aa.o - c[i]))) continue; ++ num; tmp = c[i] - d[i]; tmp = rotate(tmp, Pi / 2); tmp = itpt(p, tmp, d[i], c[i] - d[i]); len1 = dist(tmp - p); tmp = p + (tmp - p) / len1 * (R / len1); ans[num].o = (p + tmp) / 2; ans[num].r = dist(ans[num].o - p); } printf("%d\n", num); for(int i = 1; i <= num; ++i) { out(ans[i].o.fir); out(ans[i].o.sec); printf("%.8f\n", ans[i].r); } } return 0; }