人教A版高中數(shù)學(xué)(選擇性必修第一冊(cè))同步講義第21講 2.4.2圓的一般方程（含解析）

第08講2.4.2圓的一般方程課程標(biāo)準(zhǔn)學(xué)習(xí)目標(biāo)①理解與掌握?qǐng)A的一般方程的形式與條件。②能準(zhǔn)確的判定圓的存在所滿足的條件。③會(huì)判斷點(diǎn)與圓的位置關(guān)系。④會(huì)用待定系數(shù)法求圓的一般方程，并能解決與圓有關(guān)的位置、距離的綜合問題。通過本節(jié)課的學(xué)習(xí)，要求會(huì)判斷圓存在的條件，會(huì)將圓的標(biāo)準(zhǔn)形式與一般形式熟練轉(zhuǎn)化，會(huì)根椐圓存的條件求待定參數(shù)的值，會(huì)用待定系數(shù)法求圓的一般式方程，會(huì)求簡單問題中的軌跡問題，會(huì)解決與圓有關(guān)的位置與距離問題.知識(shí)點(diǎn)01：圓的一般方程對(duì)于方程SKIPIF1<0（SKIPIF1<0為常數(shù)），當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0叫做圓的一般方程.①當(dāng)SKIPIF1<0時(shí)，方程表示以SKIPIF1<0為圓心，以SKIPIF1<0為半徑的圓；②當(dāng)SKIPIF1<0時(shí)，方程表示一個(gè)點(diǎn)SKIPIF1<0③當(dāng)SKIPIF1<0時(shí)，方程不表示任何圖形說明：圓的一般式方程特點(diǎn)：①SKIPIF1<0和SKIPIF1<0前系數(shù)相等（注意相等，不一定要是1）且不為0；②沒有SKIPIF1<0項(xiàng)；③SKIPIF1<0.【即學(xué)即練1】（多選）（2022秋·高二課時(shí)練習(xí)）（多選題）下列方程不是圓的一般方程的有（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】BCD【詳解】根據(jù)二元二次方程SKIPIF1<0表示圓的條件，對(duì)于A中，方程SKIPIF1<0，可得SKIPIF1<0，所以方程是圓的一般方程；對(duì)于B中，方程SKIPIF1<0，可得SKIPIF1<0，所以方程不是圓的一般方程；對(duì)于C中，方程SKIPIF1<0中，SKIPIF1<0和SKIPIF1<0的系數(shù)不相等，所以方程不是圓的一般方程；對(duì)于D中，方程SKIPIF1<0中，存在SKIPIF1<0項(xiàng)，所以方程不是圓的一般方程.故選：BCD.知識(shí)點(diǎn)02：圓的一般方程與圓的標(biāo)準(zhǔn)方程的特點(diǎn)圓的標(biāo)準(zhǔn)方程圓的一般方程方程SKIPIF1<0SKIPIF1<0（SKIPIF1<0）圓心SKIPIF1<0SKIPIF1<0半徑SKIPIF1<0SKIPIF1<0知識(shí)點(diǎn)03：在圓的一般方程中，判斷點(diǎn)與圓的位置關(guān)系已知點(diǎn)SKIPIF1<0和圓的一般式方程SKIPIF1<0：SKIPIF1<0（SKIPIF1<0），則點(diǎn)SKIPIF1<0與圓的位置關(guān)系：①點(diǎn)SKIPIF1<0在SKIPIF1<0外SKIPIF1<0SKIPIF1<0②點(diǎn)SKIPIF1<0在SKIPIF1<0上SKIPIF1<0SKIPIF1<0③點(diǎn)SKIPIF1<0在SKIPIF1<0內(nèi)SKIPIF1<0SKIPIF1<0【即學(xué)即練2】（2022·高二課時(shí)練習(xí)）點(diǎn)SKIPIF1<0與圓SKIPIF1<0的位置關(guān)系是_____________．（填“在圓內(nèi)”、“在圓上”、“在圓外”）【答案】在圓內(nèi)【詳解】圓SKIPIF1<0的圓心坐標(biāo)為SKIPIF1<0，半徑為2點(diǎn)SKIPIF1<0到圓心的距離SKIPIF1<0，因?yàn)镾KIPIF1<0，所以點(diǎn)SKIPIF1<0在圓內(nèi)．故答案為：在圓內(nèi)題型01圓的一般方程的理解【典例1】（2022秋·安徽合肥·高二合肥市第七中學(xué)校聯(lián)考期中）已知方程SKIPIF1<0表示圓，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】因?yàn)镾KIPIF1<0表示圓，所以SKIPIF1<0，解得SKIPIF1<0，得SKIPIF1<0的取值范圍是SKIPIF1<0.故選：C【典例2】（2023·高二課時(shí)練習(xí)）方程SKIPIF1<0表示圓的充要條件是______．【答案】SKIPIF1<0或SKIPIF1<0【詳解】由題意知：SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.故答案為：SKIPIF1<0或SKIPIF1<0.【變式1】（2022秋·河南許昌·高二禹州市高級(jí)中學(xué)?？茧A段練習(xí)）方程SKIPIF1<0表示圓，則實(shí)數(shù)SKIPIF1<0的可能取值為（

）A．SKIPIF1<0 B．2 C．0 D．SKIPIF1<0【答案】D【詳解】由SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，選項(xiàng)中只有SKIPIF1<0符合題意.故選：D.【變式2】（2023春·上海寶山·高二統(tǒng)考期末）若SKIPIF1<0表示圓，則實(shí)數(shù)SKIPIF1<0的值為______.【答案】SKIPIF1<0【詳解】因?yàn)镾KIPIF1<0表示圓，所以SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)方程SKIPIF1<0，即SKIPIF1<0，不表示任何圖形，故舍去；當(dāng)SKIPIF1<0時(shí)方程SKIPIF1<0，即SKIPIF1<0，表示以SKIPIF1<0為圓心，SKIPIF1<0為半徑的圓，符合題意；故答案為：SKIPIF1<0題型02求圓的一般方程【典例1】（2023·高二課時(shí)練習(xí)）過三點(diǎn)SKIPIF1<0的圓的一般方程為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】設(shè)圓的方程為SKIPIF1<0，將A，B，C三點(diǎn)的坐標(biāo)代入方程，整理可得SKIPIF1<0，解得SKIPIF1<0，故所求的圓的一般方程為SKIPIF1<0，故選：D．【典例2】（2023·新疆克拉瑪依·高二克拉瑪依市高級(jí)中學(xué)?？计谥校┣筮m合下列條件的圓的方程：(1)圓心在直線SKIPIF1<0上，且過點(diǎn)SKIPIF1<0的圓；(2)過三點(diǎn)SKIPIF1<0的圓.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）設(shè)圓的標(biāo)準(zhǔn)方程為SKIPIF1<0，由題知：SKIPIF1<0，解得SKIPIF1<0.所以圓的標(biāo)準(zhǔn)方程為：SKIPIF1<0.（2）設(shè)圓的一般方程為：SKIPIF1<0，SKIPIF1<0，由題知：SKIPIF1<0，所以圓的方程為：SKIPIF1<0.【典例3】（2023·高二課時(shí)練習(xí)）已知圓SKIPIF1<0經(jīng)過兩點(diǎn)SKIPIF1<0，SKIPIF1<0，且圓心在直線SKIPIF1<0上，則圓SKIPIF1<0的一般方程為_______________；若直線SKIPIF1<0的方程SKIPIF1<0（SKIPIF1<0），圓心SKIPIF1<0到直線SKIPIF1<0的距離是1，則SKIPIF1<0的值是______．【答案】SKIPIF1<0SKIPIF1<0【詳解】設(shè)圓C的方程為SKIPIF1<0，由條件，得SKIPIF1<0，解得SKIPIF1<0，因此圓的一般方程為SKIPIF1<0，故圓心SKIPIF1<0，因此圓心到直線l的距離SKIPIF1<0，解得SKIPIF1<0．故答案為：SKIPIF1<0；SKIPIF1<0.【變式1】（2023·江蘇·高二假期作業(yè)）過坐標(biāo)原點(diǎn)，且在SKIPIF1<0軸和SKIPIF1<0軸上的截距分別為2和3的圓的方程為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】設(shè)圓的方程為SKIPIF1<0，由題意知,圓過點(diǎn)SKIPIF1<0，SKIPIF1<0和SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以所求圓的方程為SKIPIF1<0.故選：A【變式2】（2023·江蘇蘇州·高二蘇州中學(xué)?？计谥校┰谄矫嬷苯亲鴺?biāo)系SKIPIF1<0中，已知SKIPIF1<0的頂點(diǎn)SKIPIF1<0，SKIPIF1<0邊上中線SKIPIF1<0所在直線方程為SKIPIF1<0，SKIPIF1<0邊上的高SKIPIF1<0所在直線方程為SKIPIF1<0，求：(1)頂點(diǎn)SKIPIF1<0的坐標(biāo)；(2)SKIPIF1<0外接圓的一般方程.【答案】(1)SKIPIF1<0；(2)SKIPIF1<0.【詳解】（1）因?yàn)镾KIPIF1<0邊上的高SKIPIF1<0所在直線方程為SKIPIF1<0，所以SKIPIF1<0,解得：SKIPIF1<0.所以直線SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0.由SKIPIF1<0解得：SKIPIF1<0，即SKIPIF1<0.（2）因?yàn)辄c(diǎn)C在直線SKIPIF1<0上，所以可設(shè)SKIPIF1<0，則SKIPIF1<0中點(diǎn)為SKIPIF1<0.把SKIPIF1<0代入直線SKIPIF1<0：SKIPIF1<0，有SKIPIF1<0，解得：SKIPIF1<0，所以SKIPIF1<0.經(jīng)過SKIPIF1<0，SKIPIF1<0，SKIPIF1<0可設(shè)為：SKIPIF1<0,所以SKIPIF1<0，解得：SKIPIF1<0,所以SKIPIF1<0外接圓的方程為SKIPIF1<0.題型03圓的一般方程與標(biāo)準(zhǔn)方程轉(zhuǎn)化【典例1】（2023·高二課時(shí)練習(xí)）若圓SKIPIF1<0的圓心到直線SKIPIF1<0的距離為SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的值為（

）A．0或2 B．0或-2C．0或SKIPIF1<0 D．-2或2【答案】A【詳解】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.【典例2】（2023秋·內(nèi)蒙古巴彥淖爾·高二?？计谀┤酎c(diǎn)SKIPIF1<0為圓SKIPIF1<0的弦SKIPIF1<0的中點(diǎn)，則弦SKIPIF1<0所在直線的方程為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】圓的標(biāo)準(zhǔn)方程為SKIPIF1<0，圓心SKIPIF1<0．因?yàn)辄c(diǎn)SKIPIF1<0為弦MN的中點(diǎn)，所以SKIPIF1<0，又AP的斜率SKIPIF1<0，所以直線MN的斜率為2，弦MN所在直線的方程為SKIPIF1<0，即SKIPIF1<0.故選：D【典例3】（2023秋·高二課時(shí)練習(xí)）求圓SKIPIF1<0關(guān)于直線SKIPIF1<0的對(duì)稱圓方程.【答案】SKIPIF1<0【詳解】由SKIPIF1<0可得SKIPIF1<0，故圓心坐標(biāo)為SKIPIF1<0，半徑為1，設(shè)點(diǎn)P關(guān)于直線SKIPIF1<0的對(duì)稱點(diǎn)為SKIPIF1<0，則有SKIPIF1<0，解得SKIPIF1<0，故SKIPIF1<0，所以圓SKIPIF1<0關(guān)于直線SKIPIF1<0的對(duì)稱圓的方程為：SKIPIF1<0.【變式1】（2023春·山東青島·高二校聯(lián)考期中）圓SKIPIF1<0上的點(diǎn)到直線SKIPIF1<0的最大距離是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】圓SKIPIF1<0化為標(biāo)準(zhǔn)方程得SKIPIF1<0，圓心坐標(biāo)為SKIPIF1<0，半徑為SKIPIF1<0，圓心到直線SKIPIF1<0的距離為SKIPIF1<0所以圓上的點(diǎn)到直線SKIPIF1<0的最大距離為SKIPIF1<0.故選:C.【變式2】（2023春·遼寧朝陽·高二校聯(lián)考期中）已知點(diǎn)SKIPIF1<0在圓SKIPIF1<0上，則點(diǎn)SKIPIF1<0到SKIPIF1<0軸的距離的最大值為（

）A．2 B．3 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】圓SKIPIF1<0,即圓

SKIPIF1<0圓心為SKIPIF1<0，半徑SKIPIF1<0，得點(diǎn)P到x軸的距離的最大值為SKIPIF1<0.故選：B.題型04點(diǎn)與圓的位置關(guān)系【典例1】（2023·江蘇揚(yáng)州·高二校考階段練習(xí)）已知點(diǎn)SKIPIF1<0為圓SKIPIF1<0外一點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】因SKIPIF1<0在圓外，則SKIPIF1<0，得SKIPIF1<0.又SKIPIF1<0表示圓，則SKIPIF1<0，得SKIPIF1<0.綜上：SKIPIF1<0.故選：D【典例2】（多選）（2023·全國·高三專題練習(xí)）已知點(diǎn)SKIPIF1<0在圓SKIPIF1<0的外部，則SKIPIF1<0的取值可能是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】AC【詳解】由題意可得SKIPIF1<0，解得SKIPIF1<0，故選：AC.【變式1】（2022·高二課時(shí)練習(xí)）若點(diǎn)SKIPIF1<0是圓SKIPIF1<0內(nèi)一點(diǎn)，則過點(diǎn)SKIPIF1<0的最長的弦所在的直線方程是__________.【答案】SKIPIF1<0【詳解】圓SKIPIF1<0可整理為SKIPIF1<0，所以圓心SKIPIF1<0，SKIPIF1<0，當(dāng)過點(diǎn)SKIPIF1<0的弦經(jīng)過圓心時(shí)，弦長最長，所以過點(diǎn)SKIPIF1<0的最長的弦所在的直線方程為SKIPIF1<0，整理得SKIPIF1<0.故答案為：SKIPIF1<0.【變式2】（2023·湖北·高二校聯(lián)考期中）過點(diǎn)SKIPIF1<0可作圓SKIPIF1<0的兩條切線，則實(shí)數(shù)SKIPIF1<0的取值范圍______.【答案】SKIPIF1<0【詳解】因?yàn)榉匠蘏KIPIF1<0表示圓，過點(diǎn)SKIPIF1<0可作圓的兩條切線，則點(diǎn)SKIPIF1<0在圓外，所以SKIPIF1<0，解得：SKIPIF1<0.故答案為：SKIPIF1<0.題型05圓過定點(diǎn)問題【典例1】（2023春·上海普陀·高二曹楊二中?？茧A段練習(xí)）對(duì)任意實(shí)數(shù)SKIPIF1<0，圓SKIPIF1<0恒過定點(diǎn)，則其坐標(biāo)為______.【答案】SKIPIF1<0、SKIPIF1<0【詳解】由SKIPIF1<0由得SKIPIF1<0，故SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.故填：SKIPIF1<0、SKIPIF1<0.【典例2】（2023·高二課時(shí)練習(xí)）已知方程SKIPIF1<0表示圓，其中SKIPIF1<0，且SKIPIF1<0，則不論SKIPIF1<0取不為1的任何實(shí)數(shù)，上述圓恒過的定點(diǎn)的坐標(biāo)是________________.【答案】SKIPIF1<0【詳解】由已知得SKIPIF1<0，它表示過圓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【變式1】（2023·上海徐匯·高二上海中學(xué)?？计谥校?duì)任意實(shí)數(shù)SKIPIF1<0，圓SKIPIF1<0恒過定點(diǎn)，則定點(diǎn)坐標(biāo)為__．【答案】SKIPIF1<0或SKIPIF1<0【詳解】解：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.【變式2】（2013·遼寧大連·高二統(tǒng)考期中）對(duì)于任意實(shí)數(shù)SKIPIF1<0，曲線SKIPIF1<0恒過定點(diǎn)【答案】SKIPIF1<0【詳解】SKIPIF1<0變形為SKIPIF1<0，令SKIPIF1<0得SKIPIF1<0，所以定點(diǎn)為SKIPIF1<0故答案為：SKIPIF1<0題型06求動(dòng)點(diǎn)的軌跡方程【典例1】（2023春·上海徐匯·高二上海中學(xué)?？计谥校c(diǎn)SKIPIF1<0與兩個(gè)定點(diǎn)SKIPIF1<0，SKIPIF1<0的距離的比為SKIPIF1<0，則點(diǎn)SKIPIF1<0的軌跡方程為______．【答案】SKIPIF1<0【詳解】設(shè)點(diǎn)SKIPIF1<0，由題知SKIPIF1<0，兩邊平方化簡得SKIPIF1<0，即SKIPIF1<0，所以點(diǎn)SKIPIF1<0的軌跡方程為SKIPIF1<0.故答案為：SKIPIF1<0.【典例2】（2023·全國·高三專題練習(xí)）在平面直角坐標(biāo)系SKIPIF1<0中，曲線SKIPIF1<0與兩坐標(biāo)軸的交點(diǎn)都在圓SKIPIF1<0上.(1)求圓SKIPIF1<0的方程；(2)已知SKIPIF1<0為坐標(biāo)原點(diǎn)，點(diǎn)SKIPIF1<0在圓SKIPIF1<0上運(yùn)動(dòng)，求線段SKIPIF1<0的中點(diǎn)SKIPIF1<0的軌跡方程.【答案】(1)SKIPIF1<0(2)SKIPIF1<0（1）由SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0；令SKIPIF1<0，得SKIPIF1<0，所以圓SKIPIF1<0過SKIPIF1<0.設(shè)圓SKIPIF1<0的方程為SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，所以圓SKIPIF1<0的方程為SKIPIF1<0.（2）設(shè)SKIPIF1<0，則SKIPIF1<0，將SKIPIF1<0的坐標(biāo)代入圓SKIPIF1<0的方程得SKIPIF1<0，即SKIPIF1<0.【變式1】（2022秋·高二課時(shí)練習(xí)）過點(diǎn)SKIPIF1<0的直線與圓SKIPIF1<0交于點(diǎn)SKIPIF1<0，則線段SKIPIF1<0中點(diǎn)SKIPIF1<0的軌跡方程為___________.【答案】SKIPIF1<0【詳解】設(shè)點(diǎn)P的坐標(biāo)為SKIPIF1<0，點(diǎn)B為SKIPIF1<0，由題意，結(jié)合中點(diǎn)坐標(biāo)公式可得SKIPIF1<0，故SKIPIF1<0，化簡得SKIPIF1<0.即線段AB中點(diǎn)P的軌跡方程為SKIPIF1<0.故答案為：SKIPIF1<0【變式2】（2023春·福建莆田·高二莆田一中?？计谥校┰谄矫嬷苯亲鴺?biāo)系SKIPIF1<0中，SKIPIF1<0點(diǎn)SKIPIF1<0滿足SKIPIF1<0，則動(dòng)點(diǎn)SKIPIF1<0的運(yùn)動(dòng)軌跡方程為__________；SKIPIF1<0的最小值為__________.【答案】SKIPIF1<0SKIPIF1<0【詳解】設(shè)SKIPIF1<0，由題意可得SKIPIF1<0，整理得SKIPIF1<0，故動(dòng)點(diǎn)SKIPIF1<0的運(yùn)動(dòng)軌跡方程為SKIPIF1<0，如圖所示，點(diǎn)SKIPIF1<0的軌跡為以SKIPIF1<0為圓心，SKIPIF1<0為半徑的圓，點(diǎn)SKIPIF1<0在圓內(nèi)部，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0在線段SKIPIF1<0上時(shí)等號(hào)成立，所以SKIPIF1<0的最小值為SKIPIF1<0，故答案為：SKIPIF1<0；SKIPIF1<0題型07與圓有關(guān)的最值問題【典例1】（2023秋·北京·高二?？计谀┰O(shè)SKIPIF1<0是圓SKIPIF1<0上的動(dòng)點(diǎn)，SKIPIF1<0是圓的切線，且SKIPIF1<0，則點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0距離的最小值為（

）A．15 B．6 C．5 D．4【答案】D【詳解】解：由圓的方程SKIPIF1<0，易知圓心SKIPIF1<0，半徑為SKIPIF1<0，因?yàn)镾KIPIF1<0是圓的切線，且SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以，點(diǎn)SKIPIF1<0的軌跡方程為SKIPIF1<0，點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0距離的最小值為SKIPIF1<0，故選：D.【典例2】（2023·山東煙臺(tái)·統(tǒng)考二模）已知實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的最大值為__________．【答案】SKIPIF1<0/SKIPIF1<0【詳解】方程SKIPIF1<0整理得SKIPIF1<0，設(shè)點(diǎn)SKIPIF1<0，即點(diǎn)SKIPIF1<0是圓SKIPIF1<0上一點(diǎn)又點(diǎn)SKIPIF1<0在圓SKIPIF1<0外，所以SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0的最大值為SKIPIF1<0.故答案為：SKIPIF1<0.【典例3】（2023秋·江西宜春·高二江西省宜春市第一中學(xué)?？计谀┮阎猄KIPIF1<0為圓SKIPIF1<0上任意一點(diǎn)．則SKIPIF1<0的最大值為__________【答案】SKIPIF1<0/SKIPIF1<0【詳解】圓SKIPIF1<0即SKIPIF1<0，故圓心SKIPIF1<0，半徑為SKIPIF1<0，又SKIPIF1<0表示圓C上的點(diǎn)M到點(diǎn)SKIPIF1<0的距離，故其最大值為SKIPIF1<0，故答案為：SKIPIF1<0【變式1】（2023春·江蘇南京·高一南京市第二十九中學(xué)?？计谥校┰赟KIPIF1<0中，SKIPIF1<0，若SKIPIF1<0的平面內(nèi)有一點(diǎn)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的最小值為__________.【答案】SKIPIF1<0【詳解】由題意，由余弦定理得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，即以SKIPIF1<0為原點(diǎn)，SKIPIF1<0所在直線為SKIPIF1<0軸，SKIPIF1<0所在直線為SKIPIF1<0軸建立平面直角坐標(biāo)系，則SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，由已知SKIPIF1<0，即點(diǎn)D是在以AC的中點(diǎn)SKIPIF1<0為圓心，半徑為1的圓周上，SKIPIF1<0，即是求SKIPIF1<0的最小值，其幾何意義為圓周上的一點(diǎn)D到AB的中點(diǎn)SKIPIF1<0的距離的平方的最小值，顯然當(dāng)D，E，O共線時(shí)DE最?。ㄈ缟蠄D），即SKIPIF1<0，SKIPIF1<0的最小值為SKIPIF1<0；故答案為：SKIPIF1<0.【變式2】（2023春·江西·高二校聯(lián)考階段練習(xí)）直線SKIPIF1<0始終平分圓SKIPIF1<0的周長，則SKIPIF1<0的最小值為______．【答案】SKIPIF1<0/SKIPIF1<0【詳解】解：圓SKIPIF1<0化為標(biāo)準(zhǔn)方程：SKIPIF1<0，圓心為SKIPIF1<0，因?yàn)橹本€SKIPIF1<0始終平分圓SKIPIF1<0的周長，所以直線SKIPIF1<0過圓心SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最小值SKIPIF1<0.故答案為：SKIPIF1<0.題型08關(guān)于點(diǎn)或直線對(duì)稱的圓【典例1】（2023·全國·高三專題練習(xí)）與圓SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱的圓的標(biāo)準(zhǔn)方程是______.【答案】SKIPIF1<0【詳解】圓SKIPIF1<0的圓心SKIPIF1<0，半徑SKIPIF1<0，點(diǎn)SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱的點(diǎn)坐標(biāo)為SKIPIF1<0則所求圓的標(biāo)準(zhǔn)方程為SKIPIF1<0故答案為：SKIPIF1<0【典例2】（2023秋·重慶沙坪壩·高二重慶市第七中學(xué)校?？计谀﹫ASKIPIF1<0關(guān)于直線SKIPIF1<0的對(duì)稱圓的標(biāo)準(zhǔn)方程為__________.【答案】SKIPIF1<0【詳解】圓SKIPIF1<0的圓心SKIPIF1<0，半徑SKIPIF1<0，設(shè)點(diǎn)SKIPIF1<0關(guān)于直線SKIPIF1<0的對(duì)稱點(diǎn)SKIPIF1<0，則有SKIPIF1<0，解得SKIPIF1<0，因此所求圓的圓心SKIPIF1<0，半徑為SKIPIF1<0，所以所求圓的標(biāo)準(zhǔn)方程為：SKIPIF1<0.故答案為：SKIPIF1<0【變式1】（2023秋·山東棗莊·高二統(tǒng)考期末）如果圓SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱，則有（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】由圓的對(duì)稱性知，圓心在直線SKIPIF1<0上，故有SKIPIF1<0，即SKIPIF1<0.故選：B【變式2】（2023·江蘇·高二假期作業(yè)）已知圓SKIPIF1<0與圓SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱，則圓SKIPIF1<0的方程是__________【答案】SKIPIF1<0【詳解】圓SKIPIF1<0圓心為SKIPIF1<0，半徑等于1，設(shè)圓心SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱點(diǎn)SKIPIF1<0，則有SKIPIF1<0，且SKIPIF1<0，解得SKIPIF1<0，故點(diǎn)SKIPIF1<0，由于對(duì)稱圓SKIPIF1<0的半徑與圓SKIPIF1<0的半徑相等，故圓SKIPIF1<0的方程為SKIPIF1<0，故答案為SKIPIF1<0.題型09圓的綜合問題【典例1】（2023·全國·高三專題練習(xí)）在平面直角坐標(biāo)系SKIPIF1<0中，設(shè)二次函數(shù)SKIPIF1<0的圖象與兩坐標(biāo)軸有三個(gè)交點(diǎn)，經(jīng)過這三個(gè)交點(diǎn)的圓記為SKIPIF1<0.(1)求實(shí)數(shù)b的取值范圍；(2)求圓SKIPIF1<0的方程；(3)請(qǐng)問圓SKIPIF1<0是否經(jīng)過某定點(diǎn)（其坐標(biāo)與SKIPIF1<0無關(guān)）？請(qǐng)證明你的結(jié)論.【答案】(1)SKIPIF1<0，且SKIPIF1<0SKIPIF1<0；(2)SKIPIF1<0（SKIPIF1<0，且SKIPIF1<0）；(3)過定點(diǎn)SKIPIF1<0和SKIPIF1<0.【詳解】（1）令SKIPIF1<0得拋物線與SKIPIF1<0軸交點(diǎn)是SKIPIF1<0；令SKIPIF1<0，由題意SKIPIF1<0，且SKIPIF1<0，解得SKIPIF1<0，且SKIPIF1<0．即實(shí)數(shù)SKIPIF1<0的取值范圍SKIPIF1<0，且SKIPIF1<0SKIPIF1<0．（2）設(shè)所求圓的一般方程為SKIPIF1<0，由題意得SKIPIF1<0的圖象與兩坐標(biāo)軸的三個(gè)交點(diǎn)即為圓SKIPIF1<0和坐標(biāo)軸的交點(diǎn)，令SKIPIF1<0得，SKIPIF1<0，由題意可得，這與SKIPIF1<0是同一個(gè)方程，故SKIPIF1<0，SKIPIF1<0．令SKIPIF1<0得，SKIPIF1<0，由題意可得，此方程有一個(gè)根為SKIPIF1<0，代入此方程得出SKIPIF1<0，∴圓SKIPIF1<0的方程為SKIPIF1<0（SKIPIF1<0，且SKIPIF1<0）．（3）把圓SKIPIF1<0的方程改寫為SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，故圓SKIPIF1<0過定點(diǎn)SKIPIF1<0和SKIPIF1<0．【典例2】（2023·全國·高三專題練習(xí)）已知經(jīng)過圓SKIPIF1<0上點(diǎn)SKIPIF1<0的切線方程是SKIPIF1<0.（1）類比上述性質(zhì)，直接寫出經(jīng)過橢圓SKIPIF1<0上一點(diǎn)SKIPIF1<0的切線方程；（2）已知橢圓SKIPIF1<0，SKIPIF1<0為直線SKIPIF1<0上的動(dòng)點(diǎn)，過SKIPIF1<0作橢圓SKIPIF1<0的兩條切線，切點(diǎn)分別為SKIPIF1<0?SKIPIF1<0①求證：直線SKIPIF1<0過定點(diǎn).②當(dāng)點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離為SKIPIF1<0時(shí)，求三角形SKIPIF1<0的外接圓方程.【答案】（1）SKIPIF1<0.（2）①證明見解析；②SKIPIF1<0，SKIPIF1<0.【詳解】（1）類比上述性質(zhì)知：切線方程為SKIPIF1<0.（2）①設(shè)切點(diǎn)為SKIPIF1<0，點(diǎn)SKIPIF1<0，

由（1）的結(jié)論的AP直線方程：SKIPIF1<0，BP直線方程：SKIPIF1<0，通過點(diǎn)SKIPIF1<0，∴有SKIPIF1<0，∴A，B滿足方程：SKIPIF1<0，∴直線AB恒過點(diǎn)：SKIPIF1<0，即直線AB恒過點(diǎn)SKIPIF1<0.②已知點(diǎn)SKIPIF1<0到直線AB的距離為SKIPIF1<0.∴SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，點(diǎn)SKIPIF1<0，直線AB的方程為：SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，故點(diǎn)SKIPIF1<0.設(shè)SKIPIF1<0的外接圓方程為：SKIPIF1<0，代入得SKIPIF1<0，解得SKIPIF1<0,所以SKIPIF1<0的外接圓方程為SKIPIF1<0，即SKIPIF1<0的外接圓方程為：SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，由對(duì)稱性可知，三角形PAB的外接圓方程為：SKIPIF1<0.【變式1】（2023·全國·高二專題練習(xí)）在平面直角坐標(biāo)系SKIPIF1<0中，設(shè)二次函數(shù)SKIPIF1<0的圖象與兩坐標(biāo)軸有三個(gè)交點(diǎn)，經(jīng)過這三個(gè)交點(diǎn)的圓記為SKIPIF1<0．（Ⅰ）若SKIPIF1<0，求圓SKIPIF1<0的方程；（Ⅱ）當(dāng)SKIPIF1<0取所允許的不同的實(shí)數(shù)值時(shí)（SKIPIF1<0，且SKIPIF1<0），圓SKIPIF1<0是否經(jīng)過某定點(diǎn)（其坐標(biāo)與SKIPIF1<0無關(guān)）？請(qǐng)證明你的結(jié)論．【答案】（Ⅰ）SKIPIF1<0；（Ⅱ）SKIPIF1<0【詳解】（Ⅰ）設(shè)圓SKIPIF1<0的方程為SKIPIF1<0，令SKIPIF1<0得SKIPIF1<0，與SKIPIF1<0是同一方程，所以SKIPIF1<0，令SKIPIF1<0得SKIPIF1<0，方程SKIPIF1<0有一根為SKIPIF1<0，所以SKIPIF1<0，所以圓SKIPIF1<0的方程為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，圓C的方程為SKIPIF1<0.（Ⅱ）由（Ⅰ）知，圓SKIPIF1<0的方程為SKIPIF1<0，轉(zhuǎn)化為：SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.故圓SKIPIF1<0經(jīng)過定點(diǎn)SKIPIF1<0.【變式2】（2023秋·上海普陀·高二上海市晉元高級(jí)中學(xué)?？计谀┮阎獔AC經(jīng)過SKIPIF1<0，SKIPIF1<0兩點(diǎn)．(1)如果AB是圓C的直徑，證明：無論SKIPIF1<0取何正實(shí)數(shù)，圓SKIPIF1<0恒經(jīng)過除SKIPIF1<0外的另一個(gè)定點(diǎn)，求出這個(gè)定點(diǎn)坐標(biāo)．(2)已知點(diǎn)SKIPIF1<0關(guān)于直線SKIPIF1<0的對(duì)稱點(diǎn)SKIPIF1<0也在圓SKIPIF1<0，且過點(diǎn)SKIPIF1<0的直線SKIPIF1<0與兩坐標(biāo)軸分別交于不同兩點(diǎn)SKIPIF1<0和SKIPIF1<0，當(dāng)圓SKIPIF1<0的面積最小時(shí)，試求SKIPIF1<0的最小值.【答案】(1)證明見解析，定點(diǎn)為SKIPIF1<0(2)SKIPIF1<0【詳解】（1）設(shè)點(diǎn)SKIPIF1<0是圓SKIPIF1<0上任意一點(diǎn)，因?yàn)锳B是圓C的直徑，所以SKIPIF1<0，即SKIPIF1<0，所以圓SKIPIF1<0的方程為：SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0時(shí)等式恒成立，故定點(diǎn)為SKIPIF1<0，所以無論a取何正實(shí)數(shù)，圓C恒經(jīng)過除A外的另一個(gè)定點(diǎn)，定點(diǎn)坐標(biāo)為SKIPIF1<0；（2）因點(diǎn)A關(guān)于直線SKIPIF1<0的對(duì)稱點(diǎn)SKIPIF1<0也在圓C上，所以點(diǎn)C在直線SKIPIF1<0上，又圓C的面積最小，所以圓C是以SKIPIF1<0直徑的圓，設(shè)過點(diǎn)A與直線SKIPIF1<0垂直的直線方程為SKIPIF1<0，由方程組SKIPIF1<0得SKIPIF1<0，則SKIPIF1<0所以圓C的方程為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0或SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，由題意知直線l斜率存在且不為零，設(shè)直線l的方程為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0，時(shí)SKIPIF1<0，所以SKIPIF1<0，（當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào)）則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0題型10圓的實(shí)際應(yīng)用【典例1】（2022·高二課時(shí)練習(xí)）蘇州有很多圓拱的懸索拱橋（如寒山橋），經(jīng)測得某圓拱索橋（如圖）的跨度SKIPIF1<0米，拱高SKIPIF1<0米，在建造圓拱橋時(shí)每隔5米需用一根支柱支撐，求與SKIPIF1<0相距30米的支柱SKIPIF1<0的高度.【答案】SKIPIF1<0（米）【詳解】以SKIPIF1<0為原點(diǎn)，SKIPIF1<0所在的直線為SKIPIF1<0軸，建立如圖所示的直角坐標(biāo)系根據(jù)題意可知，SKIPIF1<0，所以SKIPIF1<0，設(shè)圓心為SKIPIF1<0，圓拱所在圓的方程為SKIPIF1<0，則因?yàn)镾KIPIF1<0在圓拱所在圓的方程上，所以SKIPIF1<0，解得SKIPIF1<0.即圓拱所在的圓方程為SKIPIF1<0，將SKIPIF1<0代入圓方程，得SKIPIF1<0，解得SKIPIF1<0SKIPIF1<0，SKIPIF1<0.所以與OP相距30米的支柱MN的高度為SKIPIF1<0（米）.【典例2】（2022秋·江西南昌·高二南昌市外國語學(xué)校校考階段練習(xí)）如圖所示，某隧道內(nèi)設(shè)雙行線公路，其截面由一段圓弧和一個(gè)長方形的三邊構(gòu)成.已知隧道總寬度SKIPIF1<0為SKIPIF1<0，行車道總寬度SKIPIF1<0為SKIPIF1<0，側(cè)墻高SKIPIF1<0，SKIPIF1<0為SKIPIF1<0，弧頂高SKIPIF1<0為SKIPIF1<0.（1）以SKIPIF1<0所在直線為SKIPIF1<0軸，SKIPIF1<0所在直線為SKIPIF1<0軸，SKIPIF1<0為單位長度建立平面直角坐標(biāo)系，求圓弧所在的圓的標(biāo)準(zhǔn)方程；（2）為保證安全，要求隧道頂部與行駛車輛頂部（設(shè)為平頂）在豎直方向上的高度之差至少為SKIPIF1<0，問車輛通過隧道的限制高度是多少？【答案】（1）SKIPIF1<0；（2）SKIPIF1<0.【詳解】（1）由題意，有SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.SKIPIF1<0所求圓的圓心在SKIPIF1<0軸上，SKIPIF1<0設(shè)圓的方程為SKIPIF1<0（SKIPIF1<0，SKIPIF1<0），SKIPIF1<0，SKIPIF1<0都在圓上，SKIPIF1<0，解得SKIPIF1<0.SKIPIF1<0圓的標(biāo)準(zhǔn)方程是SKIPIF1<0.（2）設(shè)限高為SKIPIF1<0，作SKIPIF1<0，交圓弧于點(diǎn)SKIPIF1<0，則SKIPIF1<0.將點(diǎn)SKIPIF1<0的橫坐標(biāo)SKIPIF1<0代入圓的方程，得SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0（舍去）.SKIPIF1<0.故車輛通過隧道的限制高度為SKIPIF1<0.【變式1】（2023秋·高一單元測試）如圖是一座類似于上海盧浦大橋的圓拱橋示意圖，該圓弧拱跨度SKIPIF1<0為SKIPIF1<0，圓拱的最高點(diǎn)SKIPIF1<0離水面SKIPIF1<0的高度為SKIPIF1<0，橋面SKIPIF1<0離水面SKIPIF1<0的高度為SKIPIF1<0.

(1)建立適當(dāng)?shù)钠矫嬷苯亲鴺?biāo)系，求圓拱所在圓的方程；(2)求橋面在圓拱內(nèi)部分SKIPIF1<0的長度.（結(jié)果精確到SKIPIF1<0）【答案】(1)建系見解析，圓拱方程為SKIPIF1<0，SKIPIF1<0.(2)橋面在圓拱內(nèi)部分SKIPIF1<0的長度約為367.4m【詳解】（1）設(shè)圓拱所在圓的圓心為SKIPIF1<0，以SKIPIF1<0為原點(diǎn)，SKIPIF1<0方向?yàn)镾KIPIF1<0軸正方向，SKIPIF1<0中垂線向上為SKIPIF1<0軸正方向，建立如圖所示的平面直角坐標(biāo)系.

設(shè)SKIPIF1<0與SKIPIF1<0軸交于SKIPIF1<0點(diǎn)，SKIPIF1<0與SKIPIF1<0軸交于SKIPIF1<0點(diǎn)，連接SKIPIF1<0設(shè)圓的半徑為SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，在直角SKIPIF1<0中，SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，所以圓拱方程為SKIPIF1<0，SKIPIF1<0.（2）由題意得，SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.所以橋面在圓拱內(nèi)部分SKIPIF1<0的長度約為367.4m【變式2】（2023春·上海浦東新·高二上海市實(shí)驗(yàn)學(xué)校?？计谥校┤鐖D，在寬為14的路邊安裝路燈，燈柱SKIPIF1<0高為8，燈桿SKIPIF1<0是半徑為SKIPIF1<0的圓SKIPIF1<0的一段劣弧．路燈采用錐形燈