2024年高中數(shù)學(xué)新高二暑期培優(yōu)講義第10講 圓的方程（教師版）

第第頁(yè)第10講圓的方程【題型歸納目錄】題型一：圓的標(biāo)準(zhǔn)方程題型二：圓的一般方程題型三：點(diǎn)與圓的位置關(guān)系題型四：二元二次曲線與圓的關(guān)系題型五：圓過(guò)定點(diǎn)問(wèn)題題型六：軌跡問(wèn)題【知識(shí)點(diǎn)梳理】知識(shí)點(diǎn)一：圓的標(biāo)準(zhǔn)方程SKIPIF1<0，其中SKIPIF1<0為圓心，SKIPIF1<0為半徑．知識(shí)點(diǎn)詮釋?zhuān)?1)如果圓心在坐標(biāo)原點(diǎn)，這時(shí)SKIPIF1<0，圓的方程就是．有關(guān)圖形特征與方程的轉(zhuǎn)化：如：圓心在x軸上：SKIPIF1<0；圓與y軸相切時(shí)：SKIPIF1<0；圓與x軸相切時(shí)：SKIPIF1<0；與坐標(biāo)軸相切時(shí)：SKIPIF1<0；過(guò)原點(diǎn)：SKIPIF1<0(2)圓的標(biāo)準(zhǔn)方程SKIPIF1<0圓心為SKIPIF1<0，半徑為SKIPIF1<0，它顯現(xiàn)了圓的幾何特點(diǎn)．(3)標(biāo)準(zhǔn)方程的優(yōu)點(diǎn)在于明確指出了圓心和半徑．由圓的標(biāo)準(zhǔn)方程可知，確定一個(gè)圓的方程，只需要a、b、r這三個(gè)獨(dú)立參數(shù)，因此，求圓的標(biāo)準(zhǔn)方程常用定義法和待定系數(shù)法．知識(shí)點(diǎn)二：點(diǎn)和圓的位置關(guān)系如果圓的標(biāo)準(zhǔn)方程為SKIPIF1<0，圓心為SKIPIF1<0，半徑為SKIPIF1<0，則有(1)若點(diǎn)SKIPIF1<0在圓上SKIPIF1<0(2)若點(diǎn)SKIPIF1<0在圓外SKIPIF1<0(3)若點(diǎn)SKIPIF1<0在圓內(nèi)SKIPIF1<0知識(shí)點(diǎn)三：圓的一般方程當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0叫做圓的一般方程．SKIPIF1<0為圓心，SKIPIF1<0為半徑．知識(shí)點(diǎn)詮釋?zhuān)河煞匠蘏KIPIF1<0得SKIPIF1<0(1)當(dāng)SKIPIF1<0時(shí)，方程只有實(shí)數(shù)解SKIPIF1<0．它表示一個(gè)點(diǎn)．(2)當(dāng)SKIPIF1<0時(shí)，方程沒(méi)有實(shí)數(shù)解，因而它不表示任何圖形．(3)當(dāng)SKIPIF1<0時(shí)，可以看出方程表示以SKIPIF1<0為圓心，SKIPIF1<0為半徑的圓．知識(shí)點(diǎn)四：用待定系數(shù)法求圓的方程的步驟求圓的方程常用“待定系數(shù)法”．用“待定系數(shù)法”求圓的方程的大致步驟是：(1)根據(jù)題意，選擇標(biāo)準(zhǔn)方程或一般方程．(2)根據(jù)已知條件，建立關(guān)于SKIPIF1<0或SKIPIF1<0的方程組．(3)解方程組，求出SKIPIF1<0或SKIPIF1<0的值，并把它們代入所設(shè)的方程中去，就得到所求圓的方程．知識(shí)點(diǎn)五：軌跡方程求符合某種條件的動(dòng)點(diǎn)的軌跡方程，實(shí)質(zhì)上就是利用題設(shè)中的幾何條件，通過(guò)“坐標(biāo)法”將其轉(zhuǎn)化為關(guān)于變量SKIPIF1<0之間的方程．1、當(dāng)動(dòng)點(diǎn)滿足的幾何條件易于“坐標(biāo)化”時(shí)，常采用直接法；當(dāng)動(dòng)點(diǎn)滿足的條件符合某一基本曲線的定義(如圓)時(shí)，常采用定義法；當(dāng)動(dòng)點(diǎn)隨著另一個(gè)在已知曲線上的動(dòng)點(diǎn)運(yùn)動(dòng)時(shí)，可采用代入法(或稱(chēng)相關(guān)點(diǎn)法)．2、求軌跡方程時(shí)，一要區(qū)分“軌跡”與“軌跡方程”；二要注意檢驗(yàn)，去掉不合題設(shè)條件的點(diǎn)或線等．3、求軌跡方程的步驟：(1)建立適當(dāng)?shù)闹苯亲鴺?biāo)系，用SKIPIF1<0表示軌跡(曲線)上任一點(diǎn)SKIPIF1<0的坐標(biāo)；(2)列出關(guān)于SKIPIF1<0的方程；(3)把方程化為最簡(jiǎn)形式；(4)除去方程中的瑕點(diǎn)(即不符合題意的點(diǎn))；(5)作答．【典例例題】題型一：圓的標(biāo)準(zhǔn)方程例1．圓SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱(chēng)的圓是(

)A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】圓SKIPIF1<0圓心為SKIPIF1<0，半徑為SKIPIF1<0，設(shè)點(diǎn)SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱(chēng)的點(diǎn)為SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，所以點(diǎn)SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱(chēng)的點(diǎn)為SKIPIF1<0，所以圓SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱(chēng)的圓是SKIPIF1<0.故選：D.例2．已知圓C的圓心在直線2x－y－7＝0上，且圓C與y軸交于兩點(diǎn)A(0，－4)，B(0，－2)，則圓C的標(biāo)準(zhǔn)方程為(

)A．(x－2)2＋(y－3)2＝5 B．(x－2)2＋(y＋3)2＝5C．(x＋2)2＋(y＋3)2＝5 D．(x＋2)2＋(y－3)2＝5【解析】設(shè)圓心SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，則半徑為SKIPIF1<0，圓心SKIPIF1<0.即圓C的標(biāo)準(zhǔn)方程為SKIPIF1<0.故選：B例3．已知圓SKIPIF1<0經(jīng)過(guò)點(diǎn)SKIPIF1<0，SKIPIF1<0，且圓心在直線SKIPIF1<0上，則圓SKIPIF1<0的標(biāo)準(zhǔn)方程為(

)A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】設(shè)圓SKIPIF1<0的標(biāo)準(zhǔn)方程為SKIPIF1<0，因?yàn)閳ASKIPIF1<0經(jīng)過(guò)點(diǎn)SKIPIF1<0，SKIPIF1<0，且圓心在直線SKIPIF1<0上，所以有SKIPIF1<0，因此圓SKIPIF1<0的標(biāo)準(zhǔn)方程為SKIPIF1<0，故選：A題型二：圓的一般方程例4．SKIPIF1<0三個(gè)頂點(diǎn)的坐標(biāo)分別是SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0外接圓的方程是(

)A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】設(shè)所求圓方程為SKIPIF1<0,因?yàn)镾KIPIF1<0，SKIPIF1<0，SKIPIF1<0三點(diǎn)都在圓上，所以SKIPIF1<0，解得SKIPIF1<0，即所求圓方程為：SKIPIF1<0.故選：C.例5．已知圓SKIPIF1<0經(jīng)過(guò)兩點(diǎn)SKIPIF1<0，SKIPIF1<0，且圓心SKIPIF1<0在直線SKIPIF1<0上，則圓SKIPIF1<0的方程為()A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】設(shè)圓的一般方程為SKIPIF1<0，圓心坐標(biāo)為SKIPIF1<0，因?yàn)閳ASKIPIF1<0經(jīng)過(guò)兩點(diǎn)SKIPIF1<0，SKIPIF1<0，且圓心SKIPIF1<0在直線SKIPIF1<0上，所以SKIPIF1<0，解得SKIPIF1<0，所以圓SKIPIF1<0的方程為SKIPIF1<0.故選：C.題型三：點(diǎn)與圓的位置關(guān)系例6．點(diǎn)P(m，3)與圓(x－2)2＋(y－1)2＝2的位置關(guān)系為(

)A．點(diǎn)在圓外 B．點(diǎn)在圓內(nèi) C．點(diǎn)在圓上 D．與m的值有關(guān)【解析】將點(diǎn)P(m，3)坐標(biāo)代入(x－2)2＋(y－1)2＝2中，有：SKIPIF1<0恒成立，故點(diǎn)P在圓外，故選：A.例7．點(diǎn)SKIPIF1<0與圓SKIPIF1<0的位置關(guān)系是(

)．A．在圓內(nèi) B．在圓外 C．在圓上 D．不確定【解析】因?yàn)镾KIPIF1<0，所以點(diǎn)SKIPIF1<0在圓SKIPIF1<0外.故選：B題型四：二元二次曲線與圓的關(guān)系例8．(多選題)方程SKIPIF1<0表示圓，則實(shí)數(shù)a的可能取值為(

)A．4 B．2 C．0 D．SKIPIF1<0【答案】AD【解析】把方程SKIPIF1<0整理成SKIPIF1<0，即SKIPIF1<0，若表示圓則滿足SKIPIF1<0即SKIPIF1<0，即SKIPIF1<0所以SKIPIF1<0或SKIPIF1<0，觀察答案中只有SKIPIF1<0和SKIPIF1<0符合題意.故選：AD例9．(多選題)已知方程SKIPIF1<0，下列敘述正確的是(

)A．方程表示的是圓.B．當(dāng)SKIPIF1<0時(shí)，方程表示過(guò)原點(diǎn)的圓.C．方程表示的圓的圓心在SKIPIF1<0軸上.D．方程表示的圓的圓心在SKIPIF1<0軸上.【答案】BC【解析】由SKIPIF1<0得：SKIPIF1<0；對(duì)于A，若SKIPIF1<0，即SKIPIF1<0，則方程不表示圓，A錯(cuò)誤；對(duì)于B，當(dāng)SKIPIF1<0時(shí)，方程為SKIPIF1<0，則方程表示以SKIPIF1<0為圓心，半徑為SKIPIF1<0的圓，此圓經(jīng)過(guò)原點(diǎn)，B正確；對(duì)于CD，若方程表示圓，則該圓圓心為SKIPIF1<0，半徑為SKIPIF1<0，則圓心在SKIPIF1<0軸上，不在SKIPIF1<0軸上，C正確，D錯(cuò)誤.故選：BC.題型五：圓過(guò)定點(diǎn)問(wèn)題例10．對(duì)任意實(shí)數(shù)SKIPIF1<0，圓SKIPIF1<0恒過(guò)定點(diǎn)，則定點(diǎn)坐標(biāo)為_(kāi)_．【答案】SKIPIF1<0或SKIPIF1<0【解析】SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，或SKIPIF1<0，SKIPIF1<0，所以定點(diǎn)的坐標(biāo)是SKIPIF1<0或SKIPIF1<0．故答案為：SKIPIF1<0或SKIPIF1<0.例11．對(duì)任意實(shí)數(shù)SKIPIF1<0，圓SKIPIF1<0恒過(guò)定點(diǎn)，則其坐標(biāo)為_(kāi)_____.【答案】SKIPIF1<0、SKIPIF1<0【解析】由SKIPIF1<0由得SKIPIF1<0，故SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.故填：SKIPIF1<0、SKIPIF1<0.例12．已知方程SKIPIF1<0表示圓，其中SKIPIF1<0，且a≠1，則不論a取不為1的任何實(shí)數(shù)，上述圓恒過(guò)的定點(diǎn)的坐標(biāo)是________________.【答案】SKIPIF1<0【解析】由已知得SKIPIF1<0，它表示過(guò)圓SKIPIF1<0與直線SKIPIF1<0交點(diǎn)的圓.由SKIPIF1<0，解得SKIPIF1<0即定點(diǎn)坐標(biāo)為SKIPIF1<0.故答案為SKIPIF1<0題型六：軌跡問(wèn)題例13．已知線段AB的端點(diǎn)B的坐標(biāo)為SKIPIF1<0，端點(diǎn)A在圓C：SKIPIF1<0上運(yùn)動(dòng)，求線段AB的中點(diǎn)P的軌跡方程，并說(shuō)明它的軌跡是什么．【解析】設(shè)點(diǎn)P的坐標(biāo)為SKIPIF1<0，點(diǎn)A的坐標(biāo)為SKIPIF1<0，又SKIPIF1<0，且P為線段AB的中點(diǎn)，所以SKIPIF1<0，則SKIPIF1<0.因?yàn)辄c(diǎn)A在圓C：SKIPIF1<0上運(yùn)動(dòng)，即有SKIPIF1<0，代入可得，SKIPIF1<0，整理可得SKIPIF1<0，化為標(biāo)準(zhǔn)方程可得SKIPIF1<0.所以，中點(diǎn)P的軌跡方程為SKIPIF1<0，該軌跡為以SKIPIF1<0為圓心，1為半徑的圓.例14．已知方程SKIPIF1<0表示圓，其圓心為SKIPIF1<0.(1)求圓心坐標(biāo)以及該圓半徑SKIPIF1<0的取值范圍；(2)若SKIPIF1<0，線段SKIPIF1<0的端點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，端點(diǎn)SKIPIF1<0在圓SKIPIF1<0上運(yùn)動(dòng)，求線段SKIPIF1<0中點(diǎn)SKIPIF1<0的軌跡方程.【解析】(1)方程SKIPIF1<0可變?yōu)椋篠KIPIF1<0由方程表示圓，所以SKIPIF1<0，即得SKIPIF1<0，SKIPIF1<0.圓心坐標(biāo)為SKIPIF1<0.(2)當(dāng)SKIPIF1<0時(shí)，圓SKIPIF1<0方程為：SKIPIF1<0，設(shè)SKIPIF1<0，又SKIPIF1<0為線段SKIPIF1<0的中點(diǎn)，SKIPIF1<0的坐標(biāo)為SKIPIF1<0則SKIPIF1<0，由端點(diǎn)SKIPIF1<0在圓SKIPIF1<0上運(yùn)動(dòng)，SKIPIF1<0即SKIPIF1<0SKIPIF1<0線段SKIPIF1<0中點(diǎn)SKIPIF1<0的軌跡方程為SKIPIF1<0.例15．已知圓SKIPIF1<0經(jīng)過(guò)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0三點(diǎn).(1)求圓SKIPIF1<0的方程；(2)設(shè)點(diǎn)SKIPIF1<0在圓SKIPIF1<0上運(yùn)動(dòng)，點(diǎn)SKIPIF1<0，且點(diǎn)SKIPIF1<0滿足SKIPIF1<0，記點(diǎn)SKIPIF1<0的軌跡為SKIPIF1<0，求SKIPIF1<0的方程.【解析】(1)設(shè)圓SKIPIF1<0的方程為SKIPIF1<0，將三點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0分別代入方程，則SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以圓SKIPIF1<0的方程為SKIPIF1<0；(2)設(shè)SKIPIF1<0，SKIPIF1<0，因?yàn)辄c(diǎn)SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0.因?yàn)辄c(diǎn)SKIPIF1<0在圓SKIPIF1<0上運(yùn)動(dòng)，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以點(diǎn)SKIPIF1<0的軌跡方程為SKIPIF1<0.【過(guò)關(guān)測(cè)試】一、單選題1．若圓SKIPIF1<0的圓心到直線SKIPIF1<0的距離為SKIPIF1<0，則實(shí)數(shù)a的值為(

)A．0或2 B．0或-2C．0或SKIPIF1<0 D．-2或2【答案】A【解析】將圓的方程化為標(biāo)準(zhǔn)方程為：SKIPIF1<0，所以，圓心為SKIPIF1<0，半徑SKIPIF1<0.因?yàn)閳A心SKIPIF1<0到直線的距離為SKIPIF1<0，所以，SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0.故選：A.2．已知圓SKIPIF1<0，圓SKIPIF1<0與圓SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱(chēng)，則圓SKIPIF1<0的方程為(

)A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由題意知,圓SKIPIF1<0的圓心與SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱(chēng)，且兩圓半徑相等，因?yàn)閳ASKIPIF1<0，即SKIPIF1<0，所以圓心SKIPIF1<0，半徑為SKIPIF1<0，設(shè)圓SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱(chēng)點(diǎn)為SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0，所以圓SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0.故選：A.3．若點(diǎn)SKIPIF1<0在圓SKIPIF1<0的內(nèi)部，則a的取值范圍是().A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由題可知，半徑SKIPIF1<0，所以SKIPIF1<0，把點(diǎn)SKIPIF1<0代入方程，則SKIPIF1<0，解得SKIPIF1<0，所以故a的取值范圍是SKIPIF1<0.故選：D4．動(dòng)直線SKIPIF1<0SKIPIF1<0平分圓SKIPIF1<0的周長(zhǎng)，則SKIPIF1<0的最小值(

)A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由題意，動(dòng)直線SKIPIF1<0過(guò)圓SKIPIF1<0的圓心SKIPIF1<0，則SKIPIF1<0，又SKIPIF1<0，則SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0且SKIPIF1<0，即SKIPIF1<0時(shí)，等號(hào)成立，故SKIPIF1<0的最小值為SKIPIF1<0.故選：D.二、填空題5．若l是經(jīng)過(guò)點(diǎn)SKIPIF1<0和圓SKIPIF1<0的圓心的直線，則l在y軸上的截距是________．【答案】SKIPIF1<0【解析】將圓化為標(biāo)準(zhǔn)方程可得，SKIPIF1<0，所以圓心為SKIPIF1<0.代入直線SKIPIF1<0的兩點(diǎn)式方程SKIPIF1<0，整理可得SKIPIF1<0.所以，l在y軸上的截距是SKIPIF1<0.故答案為：SKIPIF1<0.6．圓過(guò)點(diǎn)SKIPIF1<0、SKIPIF1<0，求面積最小的圓的一般方程為_(kāi)_______________．【答案】SKIPIF1<0【解析】當(dāng)SKIPIF1<0為圓的直徑時(shí)，過(guò)SKIPIF1<0、SKIPIF1<0的圓的半徑最小，從而面積最?。?yàn)辄c(diǎn)SKIPIF1<0、SKIPIF1<0，線段SKIPIF1<0的中點(diǎn)為SKIPIF1<0，SKIPIF1<0，故所求圓的半徑為SKIPIF1<0，所以，所求圓的方程為SKIPIF1<0，即SKIPIF1<0.故答案為：SKIPIF1<0.7．過(guò)圓SKIPIF1<0外一點(diǎn)SKIPIF1<0作圓的兩條切線，切點(diǎn)A、B，則SKIPIF1<0的外接圓的方程是________．【答案】SKIPIF1<0【解析】由圓SKIPIF1<0，得到圓心O坐標(biāo)為SKIPIF1<0，SKIPIF1<0,SKIPIF1<0，∴SKIPIF1<0的外接圓為四邊形SKIPIF1<0的外接圓，如圖所示，又SKIPIF1<0，∴外接圓的直徑為SKIPIF1<0，半徑為SKIPIF1<0，外接圓的圓心C為線段OP的中點(diǎn)，即SKIPIF1<0，則SKIPIF1<0的外接圓方程是SKIPIF1<0．故答案為：SKIPIF1<08．在平面直角坐標(biāo)系中，已知點(diǎn)SKIPIF1<0，點(diǎn)SKIPIF1<0在圓SKIPIF1<0上運(yùn)動(dòng)，則線段AP的中點(diǎn)SKIPIF1<0的軌跡方程是______．【答案