谁能给一些圆锥曲线系方程，如过两圆交点的圆可设为x^2+y^2+D1x+E1y+F+K(x^2+y^2+D2x+E2y+F2)=0
\u8fc7\u4e24\u5706\u4ea4\u70b9\u7684\u5706\u7cfb\u65b9\u7a0b\u5df2\u77e5\u5706A\uff1a x²+y²+D1x+E1y+F1 =0\u4e0e\u5706B\uff1ax²+y²+D2x+E2y+F2=0\uff0c
\u65b9\u7a0b\uff1ax²+y²+D1x+E1y+F1+\u03bb(x²+y²+D2x+E2y+F2)=0 \u2026\u2026 \u2460,
\u5f53\u03bb\u2260-1 \u65f6\uff0c\u65b9\u7a0b\u2460\u8868\u793a\u8fc7\u5706A\u4e0e\u5706B\u7684\u4ea4\u70b9\u7684\u5706\u7cfb\u7684\u65b9\u7a0b\uff0c\u5f53\u03bb=0\u65f6\uff0c\u8868\u793a\u5706A\uff0c\u4f46\u4e0d\u80fd\u8868\u793a\u5706B\uff1b\u5f53\u03bb=-1 \u65f6\uff0c\u82e5\u5706A\u4e0e\u5706B\u76f8\u4ea4\uff0c\u65b9\u7a0b\u2460\u8868\u793a\u5706A\u4e0e\u5706B\u7684\u516c\u5171\u5f26\u6240\u5728\u7684\u76f4\u7ebf\u65b9\u7a0b\uff0c\u5f53\u5706A\u4e0e\u5706B\u76f8\u5207\u65f6\uff0c\u65b9\u7a0b\u2460\u8868\u793a\u5706A\u4e0e\u5706B\u7684\u516c\u5207\u7ebf\u65b9\u7a0b\uff0c\u5f53\u4e24\u5706\u76f8\u79bb\u65f6\uff0c\u65b9\u7a0b\u2460\u8868\u793a\u4e0e\u4e24\u5706\u8fde\u5fc3\u7ebf\u5782\u76f4\u7684\u65b9\u7a0b\uff0c\u5728\u89e3\u5706\u7684\u6709\u5173\u95ee\u9898\uff0c\u5e38\u5e38\u7528\u5230\u8fd9\u4e00\u7ed3\u8bba\uff0c\u53ef\u4ee5\u8d77\u5230\u4e8b\u534a\u529f\u500d\u7684\u6548\u679c\u3002
\u3010\u4f8b\u3011\u6c42\u8fc7\u4e24\u5706C1:X^2+Y^2-4X+2Y=0\u548c\u5706C2:X^2+Y^2-2Y-4=0\u7684\u4ea4\u70b9\uff0c\u4e14\u5706\u5fc3\u5728\u76f4\u7ebf2x+4y-1=0\u4e0a\u7684\u5706\u7684\u65b9\u7a0b
\u3010\u89e3\u3011\u8bbe\u4e24\u5706\u4ea4\u70b9\u7684\u5706\u7cfb\u65b9\u7a0b\u4e3a\uff1a
x²+ y²-4x+2y+\u03bb(x²+y²-2y-4)=0\uff08\u4e0d\u5305\u62ecc2\uff0c\u4e14\u03bb\u2260-1\uff09
\u5373(1+\u03bb)x²+(1+\u03bb)y²-4x+2(1-\u03bb)y-4\u03bb=0
\u5706\u5fc3C\uff1a(2/(1+\u03bb),(\u03bb-1)/(1+\u03bb))
\u56e0C\u5728l\u4e0a
\u65454/(1+\u03bb)+4(\u03bb-1)/(1+\u03bb)-1=0
\u89e3\u4e4b\u03bb=1/3
\u5373C:x²+ y²-3x+y-1=0
\u5728\u65b9\u7a0b(x-a)^2+(y-b)^2=r^2\u4e2d\uff0c\u82e5\u5706\u5fc3(a,b)\u4e3a\u5b9a\u70b9\uff0cr\u4e3a\u53c2\u53d8\u6570\uff0c\u5219\u5b83\u8868\u793a\u540c\u5fc3\u5706\u7684\u5706\u7cfb\u65b9\u7a0b\uff0e\u82e5r\u662f\u5e38\u91cf\uff0ca\uff08\u6216b\uff09\u4e3a\u53c2\u53d8\u6570\uff0c\u5219\u5b83\u8868\u793a\u534a\u5f84\u76f8\u540c\uff0c\u5706\u5fc3\u5728\u540c\u4e00\u76f4\u7ebf\u4e0a\uff08\u5e73\u884c\u4e8ex\u8f74\u6216y\u8f74\uff09\u7684\u5706\u7cfb\u65b9\u7a0b\uff0e \u3000\u3000\u7ecf\u8fc7\u4e24\u5706x^2+y^2+D1x+E1y+F1=0\u4e0ex^2+y^2+D2x+E2y+F2=0 \u3000\u3000\u7684\u4ea4\u70b9\u5706\u7cfb\u65b9\u7a0b\u4e3a\uff1a \u3000\u3000x^2+y^2+D1x+E1y+F1+\u03bb(x^2+y^2+D2x+E2y+F2)=0 (\u03bb\u2260-1) \u3000\u3000\u7ecf\u8fc7\u76f4\u7ebfAx+By+C=0\u4e0e\u5706x^2+y^2+Dx+Ey+F=0\u7684\u4ea4\u70b9\u5706\u7cfb\u65b9\u7a0b \u3000\u3000x^2+y^2+Dx+Ey+F+\u03bb(Ax+By+C)=0 \u3000\u3000\u7c7b\u578b1\uff1a\u65b9\u7a0b \u8868\u793a\u534a\u5f84\u4e3a\u5b9a\u957f \u7684\u5706\u7cfb \u7c7b\u578b2\uff1a\u65b9\u7a0b \u8868\u793a\u4ee5 \u4e3a\u5706\u5fc3\u7684\u540c\u5fc3\u5706\u7cfb\u3002 \u3000\u3000\u62d3\u5c551\uff1a\u65b9\u7a0b \u8868\u793a\u5706\u5fc3\u843d\u5728\u76f4\u7ebf \u4e0a\uff0c\u534a\u5f84\u4e3a \u7684\u5706\u7cfb\u3002 \u3000\u3000\u62d3\u5c552\uff1a\u65b9\u7a0b \u8868\u793a\u5706\u5fc3\u843d\u5728\u76f4\u7ebf \u4e0a\uff0c\u534a\u5f84\u4e3a \u7684\u5706\u7cfb\u3002 \u3000\u3000\u62d3\u5c553\uff1a\u65b9\u7a0b \u8868\u793a\u5706\u5fc3\u843d\u5728\u76f4\u7ebf \u4e0a\u7684\u5706\u7cfb\u3002 \u3000\u3000\u62d3\u5c554\uff1a\u65b9\u7a0b \u8868\u793a\u5706\u5fc3\u843d\u5728\u5706 \u4e0a\uff0c\u534a\u5f84\u4e3a \u7684\u5706\u7cfb\u3002 \u3000\u3000\u7c7b\u578b3\uff1a\u5171\u8f74\u5706\u7cfb \u3000\u3000\u82e5\u2299C1\u4e0e\u2299C2\u4ea4\u4e8eA\u3001B\u4e24\u70b9\uff0c\u5219\u76f4\u7ebfAB\u79f0\u4e3a\u8fd9\u4e24\u4e2a\u5706\u7684\u6839\u8f74\u3002\u7ecf\u8fc7A\u3001B\u4e24\u70b9\u7684\u6240\u6709\u7684\u5706\u5f62\u6210\u4e00\u4e2a\u5706\u7cfb\uff0c\u8fd9\u5706\u7cfb\u5185\u4efb\u4f55\u4e24\u4e2a\u5706\u7684\u6839\u8f74\u5747\u4e3a\u76f4\u7ebfAB\uff0c\u56e0\u6b64\u6211\u4eec\u79f0\u8fd9\u79cd\u5706\u7cfb\u4e3a\u5171\u8f74\u5706\u7cfb\u3002 \u7f16\u8f91\u672c\u6bb5\u7406\u89e3\u3000\u3000\u7406\u89e3\uff1a1.\u4f8b\u9898\uff1a\u6c42x+(m+1)y+m=0\u6240\u8fc7\u5b9a\u70b9 \u3000\u3000\u89e3\uff1a\u53ef\u5c06\u539f\u5f0f\u5316\u4e3ax+y+m(y+1)=0 \u3000\u3000\u5373\u4e3ax+y=0\uff1by+1=0 \u3000\u3000\u89e3\u5f97\u6052\u8fc7\u70b9\uff081\uff0c-1\uff09 \u3000\u3000\u7531\u6b64\u6211\u4eec\u7406\u89e3\u5230\u5f53\u9664\u4e86x\uff0cy\uff08\u4e3a\u4e00\u6b21\u5e42\uff09\u8fd8\u6709\u4e00\u672a\u77e5\u6570m\u65f6,\u4f9d\u7136\u53ef\u6c42\u5f97\u4e00\u5b9a\u70b9\u3002 \u3000\u3000\u7531\u6b64\u53ef\u8054\u60f3\uff1a\u5f53\u6709\u4e8c\u6b21\u65b9\u7a0b\u7ec4x2+y2+D1x+E1y+F1=0\u4e0ex2+y2+D2x+E2y+F2=0\u6211\u4eec\u4fbf\u80fd\u6c42\u51fa\u4e24\u5b9a\u70b9\u3002 \u3000\u3000\u8fc7\u4e00\u5df2\u77e5\u5706\u4e0e\u4e00\u76f4\u7ebf\u7684\u4e24\u4e2a\u4ea4\u70b9\u7684\u5706\u7cfb\u65b9\u7a0b\u4e3a\uff1a \u3000\u3000x2+y2+D1x+E1y+F1+\u03bb\uff08Ax+By+C)=0 \u3000\u3000\u7406\u89e32\uff1a\u6709\u4e8c\u6b21\u65b9\u7a0b\u7ec4x2+y2+D1x+E1y+F1=0 \u2460\u5f0f \u3000\u3000x2+y2+D2x+E2y+F2=0 \u2461\u5f0f \u3000\u3000\u2460\u5f0f+\u2461\u5f0f\u5f97x2+y2+D1x+E1y+F1+x2+y2+D2x+E2y+F2=0 \u3000\u3000\u6b64\u65b9\u7a0b\u4ec5\u7b26\u5408\u4ea4\u70b9\u5750\u6807(\u5373\u5e26\u5165\u4ea4\u70b9\u540e\u6210\u7acb\uff09 \u3000\u3000\u52a0\u5165\u53c2\u6570\u03bb\u8ba9\u65b9\u7a0b\u4ee3\u8868\u6052\u8fc7\u4e24\u70b9\u7684\u6240\u6709\u5706\u3002 \u4f8b\u9898\u3000\u3000\u4f8b2\uff1a\u6c42\u8fc7\u4e24\u5706x2+y2=25\u548c(x-1)2+(y-1)2=16\u7684\u4ea4\u70b9\u4e14\u9762\u79ef\u6700\u5c0f\u7684\u5706\u7684\u65b9\u7a0b\u3002 \u3000\u3000\u5206\u6790\uff1a\u672c\u9898\u82e5\u5148\u8054\u7acb\u65b9\u7a0b\u6c42\u4ea4\u70b9\uff0c\u518d\u8bbe\u6240\u6c42\u5706\u65b9\u7a0b\uff0c\u5bfb\u6c42\u5404\u53d8\u91cf\u5173\u7cfb\uff0c\u6c42\u534a\u5f84\u6700\u503c\uff0c\u867d\u7136\u53ef\u884c\uff0c\u4f46\u8fd0\u7b97\u91cf\u8f83\u5927\u3002\u81ea\u7136\u9009\u7528\u8fc7\u4e24\u5706\u4ea4\u70b9\u7684\u5706\u7cfb\u65b9\u7a0b\u7b80\u4fbf\u6613\u884c\u3002\u4e3a\u4e86\u907f\u514d\u8ba8\u8bba\uff0c\u5148\u6c42\u51fa\u4e24\u5706\u516c\u5171\u5f26\u6240\u5728\u76f4\u7ebf\u65b9\u7a0b\u3002\u5219\u95ee\u9898\u53ef\u8f6c\u5316\u4e3a\u6c42\u8fc7\u4e24\u5706\u516c\u5171\u5f26\u53ca\u5706\u4ea4\u70b9\u4e14\u9762\u79ef\u6700\u5c0f\u7684\u5706\u7684\u95ee\u9898\u3002 \u3000\u3000\u89e3\uff1a\u5706x^2+y^2=25\u548c(x-1)^2+(y-1)^2=16\u7684\u516c\u5171\u5f26\u65b9\u7a0b\u4e3a \u3000\u3000x^2+y^2-25-[(x-1)^2+(y-1)^2-16]=0\uff0c\u53732x+2y-11=0 \u3000\u3000\u8fc7\u76f4\u7ebf2x+2y-11=0\u4e0e\u5706x^2+y^2=25\u7684\u4ea4\u70b9\u7684\u5706\u7cfb\u65b9\u7a0b\u4e3a \u3000\u3000x^2+y^2-25+\u03bb(2x+2y-11)=0\uff0c\u5373x^2+y^2+2\u03bby+2\u03bbx-(11\u03bb+25)=0 \u3000\u3000\u4f9d\u9898\u610f\uff0c\u6b32\u4f7f\u6240\u6c42\u5706\u9762\u79ef\u6700\u5c0f\uff0c\u53ea\u9700\u5706\u534a\u5f84\u6700\u5c0f\uff0c\u5219\u4e24\u5706\u7684\u516c\u5171\u5f26\u5fc5\u4e3a\u6240\u6c42\u5706\u7684\u76f4\u5f84\uff0c\u5706\u5fc3(-\u03bb,-\u03bb)\u5fc5\u5728\u516c\u5171\u5f26\u6240\u5728\u76f4\u7ebf2x+2y-11=0\u4e0a\u3002\u5373-2\u03bb-2\u03bb+11=0\uff0c\u5219\u03bb=-11/4 \u3000\u3000\u4ee3\u56de\u5706\u7cfb\u65b9\u7a0b\u5f97\u6240\u6c42\u5706\u65b9\u7a0b(x-11/4)^2+(y-11/4)^2=79/8 \u7f16\u8f91\u672c\u6bb5\u603b\u7ed3\u3000\u3000\u5706\u7cfb\u65b9\u7a0b\u7684\u4e3b\u8981\u667a\u6167\u662f\u5c06\u53c2\u6570\u7684\u5f62\u6001\u653e\u7f6e\u5728\u56fe\u50cf\u4e2d\u3002 \u3000\u3000\u53c2\u6570\u4e0d\u4ec5\u53ef\u5728\u4e00\u6b21\u73af\u5883\u4e2d\u8868\u793a\u4e00\u4e2a\u53d8\u91cf\uff0c\u53ef\u5728\u76f4\u89d2\u5750\u6807\u7cfb\u4e2d\u8868\u793a\u4e00\u6761\u6570\u8f74\uff0c\u8fd8\u53ef\u8ba9\u4e8c\u6b21\u56fe\u50cf\u4ee5\u4e00\u5b9a\u7684\u6761\u4ef6\u53d8\u5316\u6210\u65e0\u6570\u6761\u51fd\u6570\u56fe\u50cf\u3002
第一句话里面λ和μ没有平方。你应该拿其中的一到两个例子深入理解，举一反三，而不是花心思收集更多的。如果你真的理解了，那么你的水平就能提升。反之，如果你不理解，只是收集一堆方程，那么即使全背下来顶多也只能应付一小类题目。
再帮你补充一下：
你所谓的找规律，说白了就是猜，你只要认真证明其中一个就能明白道理了。比如你自己给的那个，我估计你只想到那个方程确是过两圆交点的圆，但是从来没想过证明所有满足要求的圆都长那个样子(必须包含K=无穷大才对)。
再给你一个方程。与椭圆x^2/m^2+y^2/n^2=1共焦的双曲线方程是x^2/(m^2+n^2-a)-y^2/a=1，其中0<a<m^2+n^2。注意，这里的常数是有范围的。别的你自己看着办。
运用曲线系解题实质上是取曲线方程中的特征量作为变量，得到曲线系，根据所给的已知量，采取待定系数法，达到解决问题的目的。常常体现的是参数变换的数学观点和整体处理的解题策略。通常的题型有求点的坐标，求曲线的方程，求图形的性质等。如下图。
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绛旓細璁捐繖涓棆杞殑瑙掑害涓何,鍙互璁＄畻鍑 cot胃 = (A-C)/B 鏂板潗鏍囩郴涓鍦嗛敟鏇茬嚎鐨勪腑蹇冧负 x'o = -(Dcos胃+Esin胃)/2 y'o = -(-Dsin胃+Ecos胃)/2 鏂版棫鍧愭爣绯荤殑鎹㈢畻鍏崇郴涓 x = x'cos胃 - y'sin胃 y = x'sin胃 + y'cos胃 鎶妜'o,y鈥榦鐨勫潗鏍囧甫鍏ュ氨鑳藉緱鍒版洸绾緼x^2+Bxy+Cy^2+...
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绛旓細2.姹鍦嗛敟鏇茬嚎鏂圭▼鐨勫父鐢ㄦ柟娉曞畾涔夋硶銆佸緟瀹氱郴鏁版硶銆佺洿鎺ユ硶銆佷唬鍏ユ硶銆佸弬鏁版硶銆佸嚑浣曟硶绛夈傚叧閿槸褰㈡暟缁撳悎,寤虹珛绛夐噺鍏崇郴銆3.瀵规湰鍗曞厓鐨勫涔犲拰鑰冭瘯瑕佹眰鑳芥牴鎹墍缁欐潯浠,閫夋嫨閫傚綋鍧愭爣绯绘眰鍑烘洸绾鏂圭▼,骞剁敾鍑烘柟绋嬫墍琛ㄧず鐨勬洸绾裤4.姹鏇茬嚎鏂圭▼鐨勪竴鑸楠ゅ強瑕佺偣鏄缓绯汇佸垪寮忋佸寲绠銆佽瘉鏄庛傜涓姝ラ鈥滃缓绯(寤虹珛鍧愭爣绯)鈥濆湪...
绛旓細C 鍒嗘瀽锛 棣栧厛鎶鍦嗛敟鏇茬嚎鏂圭▼杞寲涓虹洿瑙掑潗鏍囩郴鐨鏂圭▼锛鐒跺悗鏍规嵁鎶涚墿绾跨殑鍑嗙嚎鏂圭▼鐨勫叕寮忔眰鍑哄噯绾挎柟绋嬶紝鍐嶈浆鍖栦负鏋佸潗鏍囨柟绋嬪嵆寰楀埌绛旀锛 鍦嗛敟鏇茬嚎鐢辨瀬鍧愭爣涓庣洿瑙掑潗鏍囩郴鐨勫叧绯伙紝鍙浆鍖栦负鐩磋鍧愭爣绯讳笂鐨勬柟绋嬶紝鍗充负鎶涚墿绾縳2=8y锛屽垯鍑嗙嚎鏂圭▼涓簓=-2锛屽啀杞寲涓烘瀬鍧愭爣鏂圭▼涓合乻in胃=-2锛庢晠閫夋嫨C锛 鐐...
绛旓細鍦嗛敟鏇茬嚎,鏂圭▼鏄疉*x^2+B*y^2+a*x+b*y+C=0
