https://mathmanstudio.com/categories/%E7%9B%B4%E7%B7%9A%E8%88%87%E5%9C%93/Recent content in 直線與圓 on MathManStudio.comHugozh-TWSat, 18 Jul 2026 20:10:46 +0800https://mathmanstudio.com/posts/114%E9%AB%98%E4%B8%AD%E6%95%B8%E5%AD%B8%E5%AD%B8%E6%B8%AC%E6%95%B8a%E5%A4%9A%E9%81%B816/Sat, 18 Jul 2026 20:10:46 +0800https://mathmanstudio.com/posts/114%E9%AB%98%E4%B8%AD%E6%95%B8%E5%AD%B8%E5%AD%B8%E6%B8%AC%E6%95%B8a%E5%A4%9A%E9%81%B816/<p><img loading="lazy" src="https://mathmanstudio.com/wp-content/uploads/2026/07/%E6%88%AA%E5%9C%96-2026-07-18-%E6%99%9A%E4%B8%8A7.56.14-1024x300.png"></p> <p>這題答對率只有18%，其實比前面有些題目都還簡單，可能是大家覺得寫到後面題目會越來越難，其實不見得哦！</p> <p>自己先畫個示意圖，觀察一下應該就是用點到直線的距離和畢氏定理即可求解</p> <p>$$ L1:mx-y=3m-1 $$ $$ d(O,L1)=\frac{|1-3m|}{\sqrt{m^2+1}}=1 $$</p> <p>拆掉絕對值後可解得</p> <p>$$ m=0,\frac{3}{4}(0不合) $$</p> <p>在求原點到L2的距離</p> <p>$$ L2:mx+y=3m+1 $$ $$ d(O,L2)=\frac{|-1-3m|}{\sqrt{m^2+1}}=\frac{13}{5} $$</p> <p>利用畢氏定理 <details class="answer-box"> <summary> <span class="answer-show">查看答案</span> <span class="answer-hide">收起答案</span> </summary> <div class="answer-content"> $$ \overline{AB}=2\times\frac{12}{5}=\frac{24}{5} $$ </div> </details></p>https://mathmanstudio.com/posts/114%E9%AB%98%E4%B8%AD%E6%95%B8%E5%AD%B8%E5%AD%B8%E6%B8%AC%E6%95%B8a-%E5%A4%9A%E9%81%B88/Wed, 01 Jul 2026 20:36:41 +0800https://mathmanstudio.com/posts/114%E9%AB%98%E4%B8%AD%E6%95%B8%E5%AD%B8%E5%AD%B8%E6%B8%AC%E6%95%B8a-%E5%A4%9A%E9%81%B88/<p><img loading="lazy" src="https://mathmanstudio.com/wp-content/uploads/2026/07/%E6%88%AA%E5%9C%96-2026-07-01-%E6%99%9A%E4%B8%8A7.33.00-1024x397.png"></p> <p>將原式配方整理後可得圓方程式如下：</p> <p>$$ （x-1）^2+y^2=4 $$</p> <p>此為圓心（1,0）半徑=2之圓方程式</p> <p>$$ (1)直線方程式x=3與此圓方程式之切線，交於1點，有兩相等實根。 $$</p> <p>（2）此圖形並未於原點有點對稱的現象。</p> <p>（3）正確。</p> <p>（4）解方程式判斷有無交點即可。</p> <p>$$ y=-x+4\quad代入（x-1）^2+y^2=4 $$ $$ 整理後可得x^2-5x+\frac{13}{2}=0 $$ $$ 判別式&lt;0\rightarrow無交點 $$</p> <p>(5)求極值當然用柯西不等式啦！</p> <p>$$ 令\vec{a}=(x-1,y)\quad\vec{b}=(1,-1) $$ $$ [(x-1)^2+y^2]\times[1^2+(-1)^2]\ge（x-1-y）^2 $$ $$ 1-2\sqrt{2}\le x-y\le1+2\sqrt{2} $$</p> <p>正解：（3）（5）</p>https://mathmanstudio.com/posts/113%E9%AB%98%E4%B8%AD%E6%95%B8%E5%AD%B8%E5%88%86%E7%A7%91%E6%B8%AC%E9%A9%97%E7%B7%B4%E5%8A%9F%E5%9D%8A-%E7%9B%B4%E7%B7%9A%E8%88%87%E5%9C%93%E5%80%92%E6%95%B830%E5%A4%A9%E4%B8%8A%E6%88%B0%E5%A0%B4/Mon, 10 Jun 2024 21:37:37 +0800https://mathmanstudio.com/posts/113%E9%AB%98%E4%B8%AD%E6%95%B8%E5%AD%B8%E5%88%86%E7%A7%91%E6%B8%AC%E9%A9%97%E7%B7%B4%E5%8A%9F%E5%9D%8A-%E7%9B%B4%E7%B7%9A%E8%88%87%E5%9C%93%E5%80%92%E6%95%B830%E5%A4%A9%E4%B8%8A%E6%88%B0%E5%A0%B4/<p><img loading="lazy" src="https://mathmanstudio.com/wp-content/uploads/2024/06/7.-%E5%B9%B3%E9%9D%A2%E4%B8%8A%E6%9C%89%E4%B8%89%E9%AE%8EP-1-1-1024x162.jpg"> <img loading="lazy" src="https://mathmanstudio.com/wp-content/uploads/2024/06/X-1-856x1024.jpg"></p> <p><a href="https://youtu.be/CuP9VilZLyI">https://youtu.be/CuP9VilZLyI</a></p> <p>影片詳解</p> <p><strong>要買參考書或各種好書請至全國品牌老字號南一書局唷!點此連結購買: <a 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src="https://mathmanstudio.com/wp-content/uploads/2024/04/Red-Yellow-and-Blue-Modern-Educational-Life-of-a-Teacher-YouTube-Thumbnail-3-1024x576.png"></p>https://mathmanstudio.com/posts/113%E9%AB%98%E4%B8%AD%E6%95%B8%E5%AD%B8%E5%88%86%E7%A7%91%E6%B8%AC%E9%A9%97%E7%B7%B4%E5%8A%9F%E5%9D%8A-%E7%9B%B4%E7%B7%9A%E8%88%87%E5%9C%93-%E5%80%92%E6%95%B8121%E5%A4%A9%E4%B8%8A%E6%88%B0%E5%A0%B4/Wed, 13 Mar 2024 11:56:57 +0800https://mathmanstudio.com/posts/113%E9%AB%98%E4%B8%AD%E6%95%B8%E5%AD%B8%E5%88%86%E7%A7%91%E6%B8%AC%E9%A9%97%E7%B7%B4%E5%8A%9F%E5%9D%8A-%E7%9B%B4%E7%B7%9A%E8%88%87%E5%9C%93-%E5%80%92%E6%95%B8121%E5%A4%A9%E4%B8%8A%E6%88%B0%E5%A0%B4/<p><img loading="lazy" src="https://mathmanstudio.com/wp-content/uploads/2024/03/10.%E5%B7%B2%E7%9F%A5%E5%9C%93C%EF%BC%9A%EF%BC%88%C3%97-a%EF%BC%89-%EF%BC%88y-a%EF%BC%891a%EF%BC%9E0%EF%BC%89%E8%88%87%E7%9B%B4%E7%B7%9AL%E2%80%A6-1-1024x274.jpg"> <img loading="lazy" src="https://mathmanstudio.com/wp-content/uploads/2024/03/L_-2-x-Y-O-1-837x1024.jpg"></p> 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