圆的方程之最值问题
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【基础过关系列】2022-2023学年高二数学上学期同步知识点剖析精品讲义(人教A版2019)
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选择性必修第一册同步巩固,难度4颗星!
基础知识
最值模型
1 将军饮马
点\(A\)、\(B\)在直线\(l\)同侧,点\(P\)在直线\(l\)上,则 \((A P+B P)_{\min }=A B^{\prime}\)(当点\(A\)、\(P\)、\(B'\)共线时取到),点\(B'\)是点\(B\)关于直线\(l\)的对称点.
2 某点M到圆⊙O上点N的距离
(1) 若点\(M\)在圆内,则 \(M N_{\min }=M N_{1}=r-O M\), \(M N_{\max }=M N_{2}=r+O M\);
(2) 若点\(M\)在圆外,则 \(M N_{\min }=M N_{1}=O M-r\), \(M N_{\max }=M N_{2}=r+O M\);
3 圆上一点到圆外一定直线的距离最值
若直线\(l\)与圆\(⊙O\)相离,圆上一点\(P\)到直线\(l\)的距离为\(PE\),\(d\)为圆心\(O\)到直线\(l\)的距离,\(r\)为圆半径,则 \(P E_{\min }=P_{1} F=d-r\), \(P E_{\max }=P_{2} F=d+r\).
4 斜率型、两点距离型最值问题
形如 \(\dfrac{y-b}{x-a}\)(\(x\),\(y\)是变量,\(a\),\(b\)是常数)的式子可理解为动点\((x,y)\)与定点\((a,b)\)间的斜率;
形如 \(\sqrt{(x-a)^{2}+(y-b)^{2}}\)(\(x\),\(y\)是变量,\(a\),\(b\)是常数)的式子可理解为动点\((x,y)\)与定点\((a,b)\)间的距离.
【例】 \(\dfrac{y-2}{x+1}\)可理解为动点\((x,y)\)与定点\((-1,2)\)间的斜率;
\(\sqrt{(x+2)^{2}+y^{2}}\)可理解为动点\((x,y)\)与定点\((-2,0)\)间的距离.
圆的参数方程
圆的标准方程 \((x-a)^2+(y-b)^2=r^2\),圆心为\((a ,b)\),半径为\(r\),
它对应的圆的参数方程: \(\left\{\begin{array}{l}
x=r \cos \theta+a \\
y=r \sin \theta+b
\end{array}\right.\) (\(θ\)是参数).
解释 如图,易得\(r\cosθ=\)有向线段\(HM=x-a⇒x=r\cosθ+a\),
\(r\sinθ=\)有向线段\(HP=y-b⇒y=r\sinθ+b\).
【例】圆 \((x+1)^2+(y-2)^2=9\)的参数方程为 \(\left\{\begin{array}{l}
x=3 \cos \theta-1 \\
y=3 \sin \theta+2
\end{array}\right.\).
求最值问题的方法
对于最值问题,常用的方法有几何法(利用几何模型求解),函数法(求函数最值、三角代换),基本不等式法等.
基本方法
【题型1】圆上一点到点或直线的距离
【典题1】 在圆 \((x-2)^2+(y+3)^2=2\)上与点\((0,-5)\)距离最大的点的坐标是( )
A.\((5,1)\) \(\qquad \qquad\) B.\((4,1)\) \(\qquad \qquad\) C. \((\sqrt{2}+2, \sqrt{2}-3)\) \(\qquad \qquad\) D.\((3,-2)\)
解析 \(∵(0-2)^2+(-5+3)^2=8>2\),\(∴\)点\((0,-5)\)在圆外
\(∴\)圆上与点\((0,-5)\)距离最远的点,在圆心与点\((0,-5)\)连线上,
且与点\((0,-5)\)分别在圆心两侧令直线解析式:\(y=kx+b\),
由于直线通过点\((2,-3)\)和\((0,-5)\),可得直线解析式:\(y=x-5\),
与圆的方程联立,可得 \((x-2)^2+(x-2)^2=2\),
\(∴x=3\)或\(x=1\)
\(∴\)交点坐标为\((3,-2)\)和\((1,-4)\),其中距离点\((0,-5)\)较大的一个点为\((3,-2)\),
故选\(D\).
【典题2】 已知圆 \(x^2+(y-2)^2=1\)上一动点\(A\),定点\(B(6,1)\);\(x\)轴上一点\(W\),则\(|AW|+|BW|\)的最小值等于\(\underline{\quad \quad}\).
解析 根据题意画出圆 \(x^2+(y-2)^2=1\),以及点\(B(6,1)\)的图象如图,
作\(B\)关于\(x\)轴的对称点\(B'\),连接圆心与\(B'\),
则与圆的交点\(A\),\(|AB|\)即为\(|AW|+|BW|\)的最小值,
\(|AB|\)为点\((0,2)\)到点\(B'(6,-1)\)的距离减圆的半径,
即 \(|A B|=\sqrt{(6-0)^{2}+(-1-2)^{2}}-1=3 \sqrt{5}-1\),
故答案为: \(3 \sqrt{5}-1\).
【典题3】在平面直角坐标系\(xOy\)中,点\(Q\)为圆 \((x-1)^2+(y-1)^2=1\)上一动点,过圆\(M\)外一点\(P\)向圆\(M\)引一条切线,切点为\(A\),若\(|PA|=|PO|\),则\(|PQ|\)的最小值为( )
A. \(\sqrt{2}-1\) \(\qquad \qquad\) B. \(\sqrt{2}+1\) \(\qquad \qquad\)C. \(\dfrac{3}{4} \sqrt{2}-1\) \(\qquad \qquad\) D. \(\dfrac{3}{4} \sqrt{2}+1\)
解析 设\(P(a,b)\),
\(∵|PA|=|PO|\),圆心\(M(1,1)\),\(r=1\),
\(\therefore \sqrt{P M^{2}-r^{2}}=\sqrt{(a-1)^{2}+(b-1)^{2}-1}=\sqrt{a^{2}+b^{2}}\),
\(∴\)化简可得\(2a+2b-1=0\),
\(∵\)点\(P\)满足表达式\(2a+2b-1=0\),
\(∴\)即点\(P\)在直线\(l:2x+2y-1=0\),
由题意可知,\(|PQ|\)的最小值可转化为圆心到直线\(l\)的距离\(d\)与半径的差,
\(\therefore|P Q|=d-r=\dfrac{|1 \times 2+1 \times 2-1|}{\sqrt{2^{2}+2^{2}}}-1=\dfrac{3 \sqrt{2}}{4}-1\),
故选:\(C\).
巩固练习
1.点\(M(0,1)\)与圆 \(x^2+y^2-2x=0\)上的动点\(P\)之间的最近距离为( )
A. \(\sqrt{2}\) \(\qquad \qquad \qquad\qquad\) B.\(2\) \(\qquad\qquad \qquad \qquad\)C.\(\sqrt{2}+1\) \(\qquad \qquad \qquad\qquad\) D.\(\sqrt{2}-1\)
2.已知\(O\)为坐标原点,\(P\)为圆 \(C:(x-1)^2+(y-b)^2=1\)(常数\(b>0\))上的动点,若\(|OP|\)最大值为\(3\),则\(b\)的值为( )
A.\(1\) \(\qquad\qquad \qquad \qquad\) B.\(\sqrt{2}\) \(\qquad \qquad \qquad\qquad\) C.\(\sqrt{3}\) \(\qquad\qquad \qquad \qquad\) D.\(2\)
3.已知\(O\)为坐标原点,\(P\)为 \(⊙C:(x-a)^2+(y-1)^2=2(a>0)\)上的动点,直线\(l:x+y-1=0\),若\(P\)到\(l\)的最小距离为\(2\sqrt{2}\),则\(a\)的值为( )
A.\(2\) \(\qquad\qquad \qquad \qquad\) B.\(4\) \(\qquad \qquad \qquad\qquad\) C.\(6\) \(\qquad \qquad \qquad\qquad\) D.\(8\)
4.已知点\(A(-2,0)\),\(B(0,2)\),若点\(P\)在圆 \((x-3)^2+(y+1)^2=2\)上运动,则\(△ABP\)面积的最小值为\(\underline{\quad \quad}\).
参考答案
-
答案 \(D\)
解析 圆 \(x^2+y^2-2x=0\)可化为 \((x-1)^2+y^2=1\),
圆心为\(C(1,0)\),半径为\(r=1\);
所以 \(|M C|=\sqrt{(1-0)^{2}+(0-1)^{2}}=\sqrt{2}\),
所以点\(M\)与圆上的动点\(P\)之间的最近距离为\(|MC|-r=\sqrt{2}-1\).
故选:\(D\). -
答案 \(C\)
解析 圆 \(C:(x-1)^2+(y-b)^2=1\)的圆心为\(C(1,b)\),半径为\(1\),
所以圆\(C\)上的点\(P\)到原点的最大距离为\(|OP|=|OC|+1=3\),
即 \(\sqrt{1^{2}+b^{2}}+1=3\),解得\(b=±\sqrt{3}\),
又\(b>0\),所以\(b\)的值为\(\sqrt{3}\).
故选:\(C\). -
答案 \(C\)
解析 圆\(⊙C:(x-a)^2+(y-1)^2=2(a>0)\)的圆心坐标\(C(a,1)\),半径为\(\sqrt{2}\),
圆心到直线\(l:x+y-1=0\)的距离 \(d=\dfrac{|a+1-1|}{\sqrt{2}}\),
要使\(P\)到\(l\)的最小距离为\(2\sqrt{2}\),
则 \(\dfrac{|a+1-1|}{\sqrt{2}}=3 \sqrt{2}\),即\(|a|=6\),
又\(a>0\),\(∴a=6\).
故选:\(C\). -
答案 \(4\)
解析 \(∵\)点\(A(-2,0)\),\(B(0,2)\),若点\(P\)在圆 \((x-3)^2+(y+1)^2=2\)上运动,
\(∴AB\)的直线方程为 \(\dfrac{x}{-2}+\dfrac{y}{2}=1\),即\(x-y+2=0\).
圆心\(C(3,-1)\)到直线\(AB\)的距离为 \(d=\dfrac{|3+1+2|}{\sqrt{2}}=3 \sqrt{2}\),
则\(△ABP\)面积的最小值为 \(\dfrac{1}{2}|A B|(d-\sqrt{2})=\dfrac{1}{2} \cdot 2 \sqrt{2} \cdot 2 \sqrt{2}=4\),
故答案为:\(4\).
【题型2】斜率型、两点距离型最值问题
【典题1】 已知实数\(x\),\(y\)满足方程 \(x^2+y^2-4x+1=0\).求:
(1) \(\dfrac{y}{x}\)的最大值和最小值;
(2) \(x^2+(y+1)^2\)的最大值和最小值;
(3) \(y-x\)的最大值和最小值.
解析 (1) 方程 \(x^2+y^2-4x+1=0\)表示以点\((2,0)\)为圆心,以 \(\sqrt{3}\)为半径的圆.
\(\dfrac{y}{x}=\dfrac{y-0}{x-0}\)可理解为圆上一点\((x,y)\)与原点\((0,0)\)间的斜率,
设\(\dfrac{y}{x}=k\),即\(y=kx\),
易知圆心\((2,0)\)到\(y=kx\)的距离为半径时直线与圆相切,斜率取得最大、最小值.
由 \(\dfrac{|2 k-0|}{\sqrt{k^{2}+1}}=\sqrt{3}\),解得 \(k^2=3\).
\(∴k=\sqrt{3}\)或\(k=-\sqrt{3}\).
\(∴\dfrac{y}{x}\)的最大值为\(\sqrt{3}\),最小值为\(-\sqrt{3}\).
(2) \(x^2+(y+1)^2=(x-0)^2+(y+1)^2\)
可理解为表示圆上的一点\((x,y)\)与点\(P(0,1)\)距离的平方,
又圆心\((2,0)\)到点\(P(0,1)\)的距离为 \(\sqrt{(2-0)^{2}+(0-1)^{2}}=\sqrt{5}\),
所以 \(x^2+(y+1)^2\)的最大值是 \((\sqrt{5}+\sqrt{3})^{2}=8+2 \sqrt{15}\),
最小值是 \((\sqrt{5}-\sqrt{3})^{2}=8-2 \sqrt{15}\).
(3) 设\(b=y-x\),则\(y=x+b\),
\(b\)理解为当直线\(l:y=x+b\)与圆相交或相切时在\(y\)轴上的截距,
显然当圆与直线相切时,直线\(l\)在\(y\)轴上的截距\(b\)取到最大值与最小值,
如图直线\(l_1\)对应的\(b_1\)为最大值,直线\(l_2\)对应的\(b_2\)为最大值,
由点到直线的距离公式,得 \(\dfrac{|2-0+b|}{\sqrt{2}}=\sqrt{3}\),即\(b=-2±\sqrt{6}\).
\(∴y-x\)的最大值为\(-2+\sqrt{6}\),最小值为\(-2-\sqrt{6}\).
巩固练习
1.已知点\(P(x,y)\)在圆 \(x^2+y^2=1\)上,则 \(\sqrt{(x-1)^{2}+(y-1)^{2}}\)的最大值为\(\underline{\quad \quad}\) .
2.实数\(x\),\(y\)满足 \(x^2+y^2+2x-4y+1=0\),则 \(\dfrac{y}{x-4}\)的最大值和最小值分别为\(\underline{\quad \quad}\).
3.(多选)已知实数\(x,y\)满足方程 \(x^2+y^2-4x+1=0\),则下列说法错误的是( )
A.\(y-x\)的最大值为\(\sqrt{6}-2\)
B. \(x^2+y^2\)的最大值为\(7+4\sqrt{3}\)
C. \(\dfrac{y}{x}\)的最大值为 \(\dfrac{\sqrt{3}}{2}\)
D.\(y-x\)的最小值为\(\sqrt{6}-2\)
参考答案
-
答案 \(\sqrt{2}+1\)
解析 \(\sqrt{(x-1)^{2}+(y-1)^{2}}\)表示点\((x,y)\)与点\((1,1)\)的距离,
\(∵\)点\(P(x,y)\)在圆 \(x^2+y^2=1\)上,
\(∴\sqrt{(x-1)^{2}+(y-1)^{2}}\)的最大值为 \(\sqrt{(1+1)}+1=\sqrt{2}+1\). -
答案 最大值为\(0\),最小值为 \(-\dfrac{20}{21}\)
解析 原方程为 \((x+1)^2+(y-2)^2=4\),表示以\(P(-1,2)\) 为圆心,\(2\)为半径的圆.
设 \(k=\dfrac{y}{x-4}\),几何意义是:圆上点\(M(x,y)\)与点\(Q(4,0)\)连线的斜率.
由图可知当直线\(MQ\)是圆的切线时,\(k\)取最大值与最小值.
设切线为\(y-0=k(x-4)\),即\(kx-y-4k=0\).
圆心\(P\)到切线的距离 \(\dfrac{|-k-2-4 k|}{\sqrt{k^{2}+1}}=2\),
化简为\(21k^2+20k=0\),解得\(k=0\)或\(k=-\dfrac{20}{21}\).
\(∴ \dfrac{y}{x-4}\)的最大值为\(0\),最小值为 \(-\dfrac{20}{21}\). -
答案 \(CD\)
解析 实数\(x,y\)满足方程 \(x^2+y^2-4x+1=0\),即满足 \((x-2)^2+y^2=3\),
表示以\(C(2,0)\)为圆心,半径等于\(\sqrt{3}\)的圆.
令\(y-x=k\),即\(x-y+k=0\),当圆和直线相切时,\(k\)取得最值,
由 \(\sqrt{3}=\dfrac{|2-0+k|}{\sqrt{2}}\),求得\(k=\sqrt{6}-2\),或 \(k=-\sqrt{6}-2\),
故\(k\)的最大值为\(\sqrt{6}-2\),最小值为\(-\sqrt{6}-2\),故\(A\)正确,\(D\)错误;
由于 \(x^2+y^2\)表示圆上的点到原点距离的平方,
故它的最大值 \((2+\sqrt{3})^{2}=7+4 \sqrt{3}\),故\(B\)正确;
由于 \(\dfrac{y}{x}\)表示圆上的点与原点连线的斜率,
故当直线和圆相切时,\(\dfrac{y}{x} =m\)取得最值,
设过原点的切线方程为\(y=mx\),即\(mx-y=0\),
由 \(\sqrt{3}=\dfrac{|2 m-0|}{\sqrt{m^{2}+1}}\),求得\(m=±\sqrt{3}\),故 \(\dfrac{y}{x}\)的最大值为\(\sqrt{3}\),故\(C\)错误,
故选:\(CD\).
【题型3】三角代换
【典题1】 已知实数\(x\)、\(y\)满足方程 \(x^2+y^2-4x+1=0\).
(1)求\(y-x\)的最大值和最小值;
(2)求 \(x^2+y^2\)的最大值和最小值;
(3)求 \(\dfrac{y}{x+1}\)的取值范围.
解析 实数\(x\)、\(y\)满足方程 \(x^2+y^2-4x+1=0\),化为 \((x-2)^2+y^2=3\),
它表示一个圆,其参数方程为 \(\left\{\begin{array}{c}
x=2+\sqrt{3} \cos \alpha \\
y=\sqrt{3} \sin \alpha
\end{array}\right.\);
(1) \(y-x=\sqrt{3} \sin \alpha-(2+\sqrt{3} \cos \alpha)=\sqrt{3} \sin \alpha-\sqrt{3} \cos \alpha-2\)\(=\sqrt{6} \sin \left(\alpha-\dfrac{\pi}{4}\right)-2\)
\(\because-1 \leq \sin \left(\alpha-\dfrac{\pi}{4}\right) \leq 1\),
\(∴y-x\)的最小值为\(-2-\sqrt{6}\),最大值为\(-2+\sqrt{6}\);
(2) \(x^{2}+y^{2}=(2+\sqrt{3} \cos \alpha)^{2}+(\sqrt{3} \sin \alpha)^{2}=4 \sqrt{3} \cos \alpha+7\)
\(∴x^{2}+y^{2}\)的最大值为\(7+4\sqrt{3}\),最小值为\(7-4\sqrt{3}\);
(3) \(\dfrac{y}{x+1}=\dfrac{\sqrt{3} \sin \alpha}{3+\sqrt{3} \cos \alpha}=\dfrac{\sin \alpha}{\sqrt{3}+\cos \alpha}\)
令 \(t=\dfrac{\sin \alpha}{\sqrt{3}+\cos \alpha}\),
\(\therefore t \cos \alpha-\sin \alpha=\sqrt{3} t \Rightarrow \sqrt{t^{2}+1} \sin (\alpha-\varphi)=\sqrt{3} t\),
则 \(\sin (\alpha-\varphi)=\dfrac{\sqrt{3} t}{\sqrt{t^{2}+1}}\),
由 \(-1 \leq \dfrac{\sqrt{3} t}{\sqrt{t^{2}+1}} \leq 1\),解得 \(-\dfrac{\sqrt{2}}{2} \leq t \leq \dfrac{\sqrt{2}}{2}\),
则 \(\dfrac{y}{x+1}\)的最大值为 \(\dfrac{\sqrt{2}}{2}\),最小值为 \(-\dfrac{\sqrt{2}}{2}\).
巩固练习
1.(多选)若实数\(x,y\)满足条件 \(x^2+y^2=1\),则下列判断正确的是( )
A.\(x+y\)的范围是\([0,\sqrt{2}]\)
B.\(x^2-4x+y^2\)的范围是\([-3,5]\)
C.\(xy\)的最大值为\(1\)
D.\(\dfrac{y-2}{x+1}\)的范围是 \(\left(-\infty,-\dfrac{3}{4}\right]\)
参考答案
- 答案 \(BD\)
解析 令\(x=\cosα\),\(y=\sinα\),
则 \(x+y=\sin \alpha+\cos \alpha=\sqrt{2} \sin \left(\alpha+\dfrac{\pi}{4}\right) \in[-\sqrt{2}, \sqrt{2}]\),故\(A\)错误;
\(x^2+y^2-4x=1-4x=1-4\cosα∈[-3,5]\),故\(B\)正确;
\(x y=\sin \alpha \cos \alpha=\dfrac{1}{2} \sin 2 \alpha \in\left[-\dfrac{1}{2}, \dfrac{1}{2}\right]\),故\(C\)错误;
令 \(\dfrac{y-2}{x+1}=\dfrac{\sin \alpha-2}{\cos \alpha+1}=t\),
得 \(t \cos \alpha-\sin \alpha=-2-t\),有 \(\sqrt{t^{2}+1} \sin (\varphi-\alpha)=-2-t\),
则 \(\mid \sin (\varphi-\alpha)=\dfrac{-2-t}{\sqrt{t^{2}+1}}\),由\(\dfrac{|t+2|}{\sqrt{t^{2}+1}} \leq 1\),解得 \(t \leq-\dfrac{3}{4}\),故\(D\)正确.
故选:\(BD\).
分层练习
【A组---基础题】
1.已知\(P\)是圆 \(x^2+y^2=4\)上的动点,点\(A\)的坐标为\((0,5)\),则\(|PA|\)的最小值为( )
A.\(9\) \(\qquad \qquad\) B.\(7\)\(\qquad \qquad\) C.\(5\) \(\qquad \qquad\) D.\(3\)
2.若\(x\)、\(y\)满足 \(x^2+y^2-2x+4y-20=0\),则 \(x^2+y^2\)的最小值是( )
A.\(\sqrt{5}-5\) \(\qquad \qquad\) B.\(5-\sqrt{5}\) \(\qquad \qquad\) C.\(30-10\sqrt{5}\) \(\qquad \qquad\) D.无法确定
3.已知点\(P(7,3)\),圆 \(M:x^2+y^2-2x-10y+25=0\),点\(Q\)为在圆\(M\)上一点,点\(S\)在\(x\)轴上,则\(|SP|+|SQ|\)的最小值为( )
A.\(7\) \(\qquad \qquad\) B.\(8\) \(\qquad \qquad\) C.\(9\) \(\qquad \qquad\) D.\(10\)
4.(多选)已知点\(P(2,4)\),若过点\(Q(4,0)\)的直线\(l\)交圆 \(C:(x-6)^2+y^2=9\)于\(A\),\(B\)两点,\(R\)是圆\(C\)上动点,则( )
A.\(|AB|\)的最小值为\(2\sqrt{5}\)
B.\(P\)到\(l\)的距离的最大值为\(2\sqrt{5}\)
C.\(\overrightarrow{P Q} \cdot \overrightarrow{P R}\)的最小值为\(12-2\sqrt{5}\)
D.\(|PR|\)的最大值为\(4\sqrt{2}+3\)
5.已知直线\(l:3x+4y-1=0\),圆 \(x^2+y^2+6x+8=0\)上的点到直线\(l\)的最小距离是\(\underline{\quad \quad}\),最大距离是\(\underline{\quad \quad}\).
6.如果实数\(x,y\)满足条件: \((x-2)^2+y^2=3\),那么 \(\dfrac{y}{x}\)的最大值是\(\underline{\quad \quad}\).
7.已知圆 \(C:x^2+y^2-2y=0\),\(P\)为直线\(l:x-y-2=0\)上任一点,过点\(P\)作圆\(C\)的切线\(PT\)(\(T\)为切点),则\(|PT|\)最小值是\(\underline{\quad \quad}\).
8.(1)已知圆 \(O_1:(x-3)^2+(y-4)^2=1\),\(P(x,y)\)为圆\(O\)上的动点,求 \(d=x^2+y^2\)的最大、最小值.
(2)已知圆 \(O_2:(x+2)^2+y^2=1\),\(P(x,y)\)为圆上任一点.求 \(\dfrac{y-2}{x-1}\)最大、最小值,求\(x-2y\)的最大、最小值.
参考答案
-
答案 \(D\)
解析 圆 \(O:x^2+y^2=4\),\(∴\)圆心\(O\)为\((0,0)\),半径为\(2\);
\(∴|AO|=5\),\(∴|PA|\)的最小值为\(5-2=3\).
故选: \(D\). -
答案 \(C\)
解析 把圆的方程化为标准方程得: \((x-1)^2+(y+2)^2=25\),
则圆心\(A\)坐标为\((1,-2)\),圆的半径\(r=5\),
设圆上一点的坐标为\((x,y)\),原点\(O\)坐标为\((0,0)\),
则\(|AO|=\sqrt{5}\),\(|AB|=r=5\),
所以\(|BO|=|AB|-|OA|=5-\sqrt{5}\).
则 \(x^2+y^2\)的最小值为\((5-\sqrt{5})^2=30-10\sqrt{5}\).
故选\(C\).
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答案 \(C\)
解析 由题意知,圆的方程化为 \((x-1)^2+(y-5)^2=1\);
所以圆心\(M(1,5)\),半径为\(1\);
如图所示,作点\(P(7,3)\)关于x轴的对称点\(P'(7,-3)\);
连接\(MP'\),交圆与点\(Q\),交\(x\)轴与点\(S\),则\(|SP|+|SQ|\)的值最小;
否则,在\(x\)轴上另取一点\(S'\),连接\(S'P\),\(S'P'\),\(S'Q\),
由于\(P\)与\(P'\)关于\(x\)轴对称,所以\(|SP|=|SP'|\),\(|S'P|=|S'P'|\);
所以\(|SP|+|SQ|=|SP'|+|SQ|=|P'Q|\)\(<|S'P' |+|S'Q|=|S'P|+|S'Q|\);
(三角形中两边之和大于第三边).
故\(|SP|+|SQ|\)的最小值为 \(\left|P^{\prime} M\right|-1=\sqrt{(1-7)^{2}+(5+3)^{2}}-1=9\);
故选:\(C\).
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答案 \(ABD\)
解析 如图,当直线\(l\)与\(x\)轴垂直时,\(|AB|\)有最小值,且最小值为\(2\sqrt{5}\),故\(A\)正确;
当直线\(l\)与\(PQ\)垂直时,\(P\)到\(l\)的距离有最大值,且最大值为\(|PQ|=2\sqrt{5}\),故\(B\)正确;
设\(R(6+3\cosθ,3\sinθ)\),
则 \(\overrightarrow{P Q} \cdot \overrightarrow{P R}=(2,-4) \cdot(4+3 \cos \theta, 3 \sin \theta-4)\)\(=6 \cos \theta-12 \sin \theta+24\),
\(\therefore \overrightarrow{P Q} \cdot \overrightarrow{P R}=6 \sqrt{5} \cos (\theta+\varphi)+24\),
则 \(\overrightarrow{P Q} \cdot \overrightarrow{P R}\)的最小值为\(24-6\sqrt{5}\),故\(C\)错误;
当\(P\),\(C\),\(R\)三点共线时,\(|PR|\)最大,且最大值为\(|PC|+r=4\sqrt{2}+3\),所以\(D\)正确.
故选:\(ABD\).
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答案 最小距离为1,最大距离为3.
解析 圆心到直线的距离加、减圆的半径,就是所求的最大值与最小值.
\(∵\)圆的方程为 \((x+3)^2+y^2=1\),
\(\therefore \dfrac{|3 \times(-3)+4 \times 0-1|}{\sqrt{3^{2}+4^{2}}} \pm 1=2 \pm 1\).
\(∴\)最小距离为\(1\),最大距离为\(3\). -
答案 \(\sqrt{3}\)
解析 满足等式 \((x-2)^2+y^2=3\)的图形如图所示:
\(\dfrac{y}{x}\)表示圆上动点与原点\(O\)连线的斜率,
由图可得动点与\(B\)重合时,此时\(OB\)与圆相切, \(\dfrac{y}{x}\)取最大值,
连接\(BC\),在\(Rt△OBC\)中,\(BC=\sqrt{3}\),\(OC=2\)
\(\because \sin \angle B O C=\dfrac{B C}{O C}=\dfrac{\sqrt{3}}{2}\),
\(∴∠BOC=60°\),此时\(\dfrac{y}{x}=\sqrt{3}\).
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答案 \(\dfrac{\sqrt{14}}{2}\)
解析 圆 \(C:x^2+(y-1)^2=1\),圆心\(C(0,1)\),半径\(r=1\),
设圆心\(C\)到直线\(l:x-y-2=0\)的距离为\(d\),
故当圆心\(C\)到直线\(l\)上点的距离最小时,
即圆心到直线的距离\(d\),此时\(|PT|\)最小,
因为 \(d=\dfrac{|-1-2|}{\sqrt{2}}=\dfrac{3 \sqrt{2}}{2}\),
所以 \(P T=\sqrt{d^{2}-r^{2}}=\sqrt{\left(\dfrac{3 \sqrt{2}}{2}\right)^{2}-1}=\dfrac{\sqrt{14}}{2}\),
故\(|PT|\)最小值是 \(\dfrac{\sqrt{14}}{2}\). -
答案 (1) \(d_{\max }=36, d_{\min }=16\)
(2)\(\dfrac{y-2}{x-1}\)的 最大值为 \(\dfrac{3+\sqrt{3}}{4}\),最小值为 \(\dfrac{3-\sqrt{3}}{4}\);
\(x-2y\)的最大值为\(-2+\sqrt{5}\),最小值为\(-2-\sqrt{5}\).
解析 (1)(法1)由圆的标准方程\((x-3)^2+(y-4)^2=1\).
可设圆的参数方程为 \(\left\{\begin{array}{l} x=3+\cos \theta \\ y=4+\sin \theta \end{array}\right.\)(\(θ\)是参数).
则 \(d=x^{2}+y^{2}=(3+\cos \theta)^{2}+(4+\sin \theta)^{2}\)
\(=26+6 \cos \theta+8 \sin \theta=26+10 \sin (\theta+\varphi)\)(其中 \(\tan \theta=\dfrac{3}{4}\))
所以 \(d_{\max }=36, d_{\min }=16\).
(法2)圆上点到原点距离的最大值\(d_1\)等于圆心到原点的距离加上半径\(1\),
圆上点到原点距离的最小值\(d_2\)等于圆心到原点的距离减去半径\(1\).
所以 \(d_{1}=\sqrt{3^{2}+4^{2}}+1=6\), \(d_{2}=\sqrt{3^{2}+4^{2}}-1=4\).
所以 \(d_{\max }=36, d_{\min }=16\).
(2) (法1)由 \((x+2)^2+y^2=1\)
得圆的参数方程:\(\left\{\begin{array}{c} x=-2+\cos \theta \\ y=\sin \theta \end{array}\right.\)(\(θ\)是参数).
则 \(\dfrac{y-2}{x-1}=\dfrac{\sin \theta-2}{\cos \theta-3}\),
令 \(t=\dfrac{\sin \theta-2}{\cos \theta-3}\),得 \(\sin \theta-t \cos \theta=2-3 t\),
\(\sqrt{1+t^{2}} \sin (\theta-\varphi)=2-3 t \Rightarrow\left|\dfrac{2-3 t}{\sqrt{1+t^{2}}}\right|=|\sin (\theta-\varphi)| \leq 1\)
\(\Rightarrow \dfrac{3-\sqrt{3}}{4} \leq t \leq \dfrac{3+\sqrt{3}}{4}\).
所以 \(t_{\max }=\dfrac{3+\sqrt{3}}{4}\), \(t_{\min }=\dfrac{3-\sqrt{3}}{4}\),
即 \(\dfrac{y-2}{x-1}\)的最大值为 \(\dfrac{3+\sqrt{3}}{4}\),最小值为 \(\dfrac{3-\sqrt{3}}{4}\).
此时\(x-2y=-2+\cosθ-2\sinθ=-2+√5 \sin(θ+φ)\).
所以\(x-2y\)的最大值为\(-2+\sqrt{5}\),最小值为\(-2-\sqrt{5}\).
(法2)设 \(\dfrac{y-2}{x-1}=k\),则\(kx-y-k+2=0\).
由于\(P(x,y)\)是圆上点,当直线与圆有交点时,如图所示,
两条切线的斜率分别是最大、最小值.
由 \(d=\dfrac{|-2 k-k+2|}{\sqrt{1+k^{2}}}=1\),得 \(k=\dfrac{3 \pm \sqrt{3}}{4}\).
所以\(\dfrac{y-2}{x-1}\)的 最大值为 \(\dfrac{3+\sqrt{3}}{4}\),最小值为 \(\dfrac{3-\sqrt{3}}{4}\).
令\(x-2y=t\),同理两条切线在\(x\)轴上的截距分别是最大、最小值.
由 \(d=\dfrac{|-2-m|}{\sqrt{5}}=1\),得\(m=-2±\sqrt{5}\).
所以\(x-2y\)的最大值为\(-2+\sqrt{5}\),最小值为\(-2-\sqrt{5}\).
【B组---提高题】
1.已知 \(⊙C:x^2+y^2-2x-2y-2=0\),直线\(l:x+2y+2=0\),\(M\)为直线\(l\)上的动点,过点\(M\)作\(⊙C\)的切线\(MA\),\(MB\),切点为\(A\),\(B\),当四边形\(MACB\)的面积取最小值时,直线\(AB\)的方程为( )
A.\(x+2y-1=0\) \(\qquad \qquad\) B.\(x+2y+1=0\) \(\qquad \qquad\) C.\(x-2y-1=0\) \(\qquad \qquad\) D.\(x-2y+1=0\)
2.已知直线\(l:y=x+3\)与\(x\)轴的交点为\(A(-3,0)\),\(P\)是直线\(l\)上任一点,过点\(P\)作圆 \(E:(x-1)^2+y^2=4\)的两条切线,设切点分别为\(C\)、\(D\),\(M\)是线段\(CD\)的中点,则\(|AM|\)的最大值为( )
A.\(2\sqrt{2}\) \(\qquad \qquad\) B.\(3\sqrt{2}\) \(\qquad \qquad\) C. \(\dfrac{7 \sqrt{2}}{2}\) \(\qquad \qquad\) D.\(4\sqrt{2}\)
参考答案
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答案 \(B\)
解析 \(⊙C:x^2+y^2-2x-2y-2=0\)的标准方程为 \((x-1)^2+(y-1)^2=4\),
则圆心\(C(1,1)\),半径\(r=2\).
因为四边形\(MACB\)的面积 \(S=2 S_{\triangle C A M}= | C A|\cdot| A M|=2| A M \mid=2 \sqrt{|C M|^{2}-4}\),
要使四边形\(MACB\)面积最小,则需\(|CM|\)最小,此时\(CM\)与直线l垂直,
直线\(CM\)的方程为\(y-1=2(x-1)\),即\(y=2x-1\),
联立 \(\left\{\begin{array}{l} y=2 x-1 \\ x+2 y+2=0 \end{array}\right.\),解得\(M(0,-1)\).则\(|CM|=\sqrt{5}\)
则以\(CM\)为直径的圆的方程为 \(\left(x-\dfrac{1}{2}\right)^{2}+y^{2}=\dfrac{5}{4}\),
与\(⊙C\)的方程作差可得直线\(AB\)的方程为\(x+2y+1=0\).
故选:\(B\).
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答案 \(B\)
解析 由题意得\(E(1,0)\),半径\(r=2\),设\(P(x_0,y_0)\),
以\(EP\)为直径的圆的方程为 \(x^2+y^2-(1+x_0 )x-y_0 y+x_0=0\),
由题意得\(ED⊥DP\),\(EC⊥CP\),
所以\(C\),\(D\)在以\(EP\)为直径的圆上,
联立 \(\left\{\begin{array}{l} x^{2}+y^{2}-\left(1+x_{0}\right) x-y_{0} y+x_{0}=0 \\ (x-1)^{2}+y^{2}=4 \end{array}\right.\),
两个方程相减得\((1-x)(1-x_0)+y_0 y=4\)为直线\(CD\)的方程,
因为\(M\)为\(CD\)的中点,
所以在以\(EP\)为直径的圆中,\(M\)为弦\(CD\)的中点,
所以\(EM⊥CD\),
所以直线\(EM\)的方程为 \(y=\dfrac{-y_{0}}{2-x_{0}}(x-1)\),
因为\(P\)在\(l\)上,故\(y_0=x_0+3\),
联立 \(\left\{\begin{array}{l} \left(2-x_{0}\right) x-y_{0} y+3=0 \\ y=\dfrac{-y_{0}}{2-x_{0}}(x-1) \\ y_{0}=3+x_{0} \end{array}\right.\),
得\(\left(x-\dfrac{1}{2}\right)^{2}+\left(y-\dfrac{1}{2}\right)^{2}=\dfrac{1}{2}\)
所以\(M\)的轨迹是以 \(\left(\dfrac{1}{2}, \dfrac{1}{2}\right)\)为圆心,以 \(\dfrac{\sqrt{2}}{2}\)为半径的圆,
点\(A(-3,0)\)到圆心的距离 \(d=\sqrt{\left(-3-\dfrac{1}{2}\right)^{2}+\left(\dfrac{1}{2}\right)^{2}}=\dfrac{5 \sqrt{2}}{2}\),
故 \(\dfrac{5 \sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}=3 \sqrt{2}\).
故选:\(B\).
【C组---拓展题】
1.已知点\(P,Q\)分别在直线\(l_1:x+y+2=0\)与直线\(l_2:x+y-1=0\)上,且\(PQ⊥l_1\),点\(A(-3,-3)\), \(B\left(\dfrac{3}{2}, \dfrac{1}{2}\right)\),则\(|AP|+|PQ|+|QB|\)的最小值为( )
A. \(\dfrac{\sqrt{130}}{2}\) \(\qquad \qquad\) B. \(\sqrt{13}+\dfrac{3 \sqrt{2}}{2}\) \(\qquad \qquad\) C.\(\sqrt{13}\) \(\qquad \qquad\) D.\(3\sqrt{2}\)
2.如图,在平面直角坐标系\(xOy\)中,正方形\(OABC\)的边长为\(4\),\(E(0,1)\),点\(F\)是正方形边\(OC\)上的一个动点,点\(O\)关于直线\(EF\)的对称点为\(G\)点,当 \(|\overrightarrow{G A}+3 \overrightarrow{G B}|\)取得最小值时,直线\(GF\)的方程为\(\underline{\quad \quad}\).
参考答案
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答案 \(B\)
解析 由平行线距离公式得: \(|P Q|=\dfrac{3}{\sqrt{2}}=\dfrac{3 \sqrt{2}}{2}\),
设\(P(a,-a-2)\),则 \(Q\left(a+\dfrac{3}{2},-a-\dfrac{1}{2}\right)\),
所以 \(|A P|+|P Q|+|Q B|\)\(=\sqrt{(a+3)^{2}+(-a+1)^{2}}+\sqrt{a^{2}+(-a-1)^{2}}+\dfrac{3 \sqrt{2}}{2}\)
\(=\sqrt{(a+3)^{2}+(a-1)^{2}}+\sqrt{a^{2}+(a+1)^{2}}+\dfrac{3 \sqrt{2}}{2}\),
设点\(M(a,a)\),\(C(1,-3)\),\(D(-1,0)\),如下图:
则有 \(\sqrt{(a+3)^{2}+(a-1)^{2}}+\sqrt{a^{2}+(a+1)^{2}}=|M C|+|M D|\)\(\geq|C D|=\sqrt{13}\)
(即当\(D\)、\(M\)、\(C\)三点共线时等号成立),
综上, \(|A P|+|P Q|+|Q B| \geq \sqrt{13}+\dfrac{3 \sqrt{2}}{2}\).
故选:\(B\). -
答案 \(y=-x+1+\sqrt{2}\)
解析 方法一 函数法
设\(F(t,0)\),\((0<t≤4)\).
则直线\(EF\)的方程为: \(\dfrac{x}{t}+y=1\),可得 \(y=-\dfrac{x}{t}+1\).
设\(G(a,b)\),\((a,b>0)\).
则 \(\dfrac{b}{a} \times\left(-\dfrac{1}{t}\right)=-1\), \(\dfrac{a}{2 t}+\dfrac{b}{2}=1\),
联立解得 \(a=\dfrac{2 t}{1+t^{2}}\), \(b=\dfrac{2 t^{2}}{1+t^{2}}\).
\(\therefore G\left(\dfrac{2 t}{1+t^{2}}, \dfrac{2 t^{2}}{1+t^{2}}\right)\).
\(\because \overrightarrow{G A}+3 \overrightarrow{G B}=(-a, 4-b)+3(4-a, 4-b)=(12-4 a, 16-4 b)\).
\(\therefore|\overrightarrow{G A}+3 \overrightarrow{G B}|=\sqrt{(12-4 a)^{2}+(16-4 b)^{2}}=2 \sqrt{(3-a)^{2}+(4-b)^{2}}\) \(=2 \sqrt{\left(3-\dfrac{2 t}{1+t^{2}}\right)^{2}+\left(4-\dfrac{2 t^{2}}{1+t^{2}}\right)^{2}}=2 \sqrt{13-12 \cdot \dfrac{t-1}{1+t^{2}}}\),
令\(m=t-1∈(-1,3]\),
\(|\overrightarrow{G A}+3 \overrightarrow{G B}|=2 \sqrt{13-12 \cdot \dfrac{m}{m^{2}+2 m+2}}\)
易得 \(f(m)=\dfrac{m}{m^{2}+2 m+2}=\left\{\begin{array}{cc} 0 & m=0 \\ \dfrac{1}{m+\dfrac{1}{m}+2}, & m \in(-1,0) \bigcup(0,3] \end{array}\right.\)的最大值在 \(m=\sqrt{2}\),
即\(t=\sqrt{2}+1\)时取到,
此时 \(G\left(\dfrac{\sqrt{2}}{2}, 1+\dfrac{\sqrt{2}}{2}\right)\), \(F(\sqrt{2}+1,0)\),
易得直线\(GF\)方程为\(y=-x+1+\sqrt{2}\).
方法二 几何法
设点\(G(a,b)\),
\(∵\)点\(O\)关于直线\(EF\)的对称点为\(G\)点,
\(∴EG=OE=1\),
\(∴\)点\(G\)的轨迹是以\(E\)为圆心,半径为\(1\)的圆 \(\odot E: a^{2}+(b-1)^{2}=1\),
\(\because \overrightarrow{G A}+3 \overrightarrow{G B}=(-a, 4-b)+3(4-a, 4-b)=(12-4 a, 16-4 b)\),
\(\therefore|\overrightarrow{G A}+3 \overrightarrow{G B}|=\sqrt{(12-4 a)^{2}+(16-4 b)^{2}}=2 \sqrt{(3-a)^{2}+(4-b)^{2}}\),
设点\(P(3,4)\),
则\(|\overrightarrow{G A}+3 \overrightarrow{G B}|\)取到最小值时,即点\(P\)到圆\(\odot E\)上一点最短距离之时,
此时点\(G\)为直线\(EP:y=x+1\)与圆 \(\odot E\)的交点,
由 \(\left\{\begin{array}{c} y=x+1 \\ x^{2}+(y-1)^{2}=1 \end{array}\right.\)
解得 \(x=\dfrac{\sqrt{2}}{2}\), \(y=1+\dfrac{\sqrt{2}}{2}\),即 \(G\left(\dfrac{\sqrt{2}}{2}, 1+\dfrac{\sqrt{2}}{2}\right)\),
此时\(GF⊥EP\),故直线\(GF\)的斜率为\(-1\),
故直线\(GF\)方程为\(y=-x+1+\sqrt{2}\).
方法三 三角代换
如方法二,得到\(a^2+(b-1)^2 =1\),
可设\(a=\sinα\),\(b=1+\cosα\),
则 \(|\overrightarrow{G A}+3 \overrightarrow{G B}|=\sqrt{(12-4 a)^{2}+(16-4 b)^{2}}\)
\(=2 \sqrt{(3-\sin \alpha)^{2}+(3-\cos \alpha)^{2}}\)
\(=2 \sqrt{19-6 \sqrt{2} \sin \left(\alpha+\dfrac{\pi}{4}\right)}\),
当 \(\alpha=\dfrac{\pi}{4}\),即 \(G\left(\dfrac{\sqrt{2}}{2}, 1+\dfrac{\sqrt{2}}{2}\right)\)时, \(|\overrightarrow{G A}+3 \overrightarrow{G B}|\)取到最小值.
设点\(F(t,0)\),
由\(GF=OF\)得 \(\sqrt{\left(t-\dfrac{\sqrt{2}}{2}\right)^{2}+\left(1+\dfrac{\sqrt{2}}{2}\right)^{2}}=t\),解得\(t=\sqrt{2}+1\),
易得直线\(GF\)方程为\(y=-x+1+\sqrt{2}\).