What-is-the-equation-of-a-circle-that-goes-through-points-0-1-1-4-and-5-2-

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Question Number 134772 by bramlexs22 last updated on 07/Mar/21

$$ \\ $$What is the equation of a circle that goes through points (0,1), (1,4), and (5,2)?

Commented by bramlexs22 last updated on 07/Mar/21

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{all}\:\mathrm{sir}\:\mathrm{and}\:\mathrm{master} \\ $$

Answered by EDWIN88 last updated on 07/Mar/21

The perpendicular bisector of any chord of a circle is a diameter and thus goes through the center of said circle Midpoint of { ((P_1 (0,1))),((P_2 (1,4))) :} is S((1/2),(5/2)) and the slope is m = ((4−1)/(1−0)) = 3, the perpendicular slope is m_p =−(1/3). the line equation ⇒y=−(1/3)(x−(1/2))+(5/2) ⇒y=−(1/3)x+(8/3) Midpoint of { ((P_2 (1,4))),((P_3 (5,2))) :}is Q(3,3) and the slope is m_2 =((4−2)/(1−5))=−(1/2),the perpendicular slope is m_p_2 = 2 , the line equation ⇒y=2(x−3)+3 ; y = 2x−3 set the two equation to find x and y, after solving we get { ((x= ((17)/7))),((y=((13)/7))) :} , so the center point is (((17)/7),((13)/7)) with radius =(√((0−((17)/7))^2 +(1−((13)/7))^2 )) r =(√((289+36)/(49))) = (√((325)/(49))) Therefore the circle′s equation is (x−((17)/7))^2 +(y−((13)/7))^2 = ((325)/(49))

$$\mathrm{The}\:\mathrm{perpendicular}\:\mathrm{bisector}\:\mathrm{of}\:\mathrm{any}\:\mathrm{chord}\: \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{is}\:\mathrm{a}\:\mathrm{diameter}\:\mathrm{and}\:\mathrm{thus}\:\mathrm{goes}\:\mathrm{through} \\ $$$$\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:\mathrm{said}\:\mathrm{circle}\: \\ $$$$\mathrm{Midpoint}\:\mathrm{of}\:\begin{cases}{\mathrm{P}_{\mathrm{1}} \left(\mathrm{0},\mathrm{1}\right)}\\{\mathrm{P}_{\mathrm{2}} \left(\mathrm{1},\mathrm{4}\right)}\end{cases}\:\mathrm{is}\:\mathrm{S}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{5}}{\mathrm{2}}\right) \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{slope}\:\mathrm{is}\:\mathrm{m}\:=\:\frac{\mathrm{4}−\mathrm{1}}{\mathrm{1}−\mathrm{0}}\:=\:\mathrm{3},\:\mathrm{the}\:\mathrm{perpendicular} \\ $$$$\mathrm{slope}\:\mathrm{is}\:{m}_{{p}} \:=−\frac{\mathrm{1}}{\mathrm{3}}.\:\mathrm{the}\:\mathrm{line}\:\mathrm{equation}\: \\ $$$$\Rightarrow\mathrm{y}=−\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{y}=−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}+\frac{\mathrm{8}}{\mathrm{3}} \\ $$$$\mathrm{Midpoint}\:\mathrm{of}\:\begin{cases}{\mathrm{P}_{\mathrm{2}} \left(\mathrm{1},\mathrm{4}\right)}\\{\mathrm{P}_{\mathrm{3}} \left(\mathrm{5},\mathrm{2}\right)}\end{cases}\mathrm{is}\:\mathrm{Q}\left(\mathrm{3},\mathrm{3}\right) \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{slope}\:\mathrm{is}\:\mathrm{m}_{\mathrm{2}} =\frac{\mathrm{4}−\mathrm{2}}{\mathrm{1}−\mathrm{5}}=−\frac{\mathrm{1}}{\mathrm{2}},\mathrm{the}\:\mathrm{perpendicular} \\ $$$$\mathrm{slope}\:\mathrm{is}\:{m}_{{p}_{\mathrm{2}} } =\:\mathrm{2}\:,\:\mathrm{the}\:\mathrm{line}\:\mathrm{equation}\: \\ $$$$\Rightarrow\mathrm{y}=\mathrm{2}\left(\mathrm{x}−\mathrm{3}\right)+\mathrm{3}\:;\:\mathrm{y}\:=\:\mathrm{2x}−\mathrm{3} \\ $$$$\mathrm{set}\:\mathrm{the}\:\mathrm{two}\:\mathrm{equation}\:\mathrm{to}\:\mathrm{find}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y},\:\mathrm{after} \\ $$$$\mathrm{solving}\:\mathrm{we}\:\mathrm{get}\:\begin{cases}{\mathrm{x}=\:\frac{\mathrm{17}}{\mathrm{7}}}\\{\mathrm{y}=\frac{\mathrm{13}}{\mathrm{7}}}\end{cases}\:,\:\mathrm{so}\:\mathrm{the}\:\mathrm{center} \\ $$$$\mathrm{point}\:\mathrm{is}\:\left(\frac{\mathrm{17}}{\mathrm{7}},\frac{\mathrm{13}}{\mathrm{7}}\right)\:\mathrm{with}\:\mathrm{radius}\:=\sqrt{\left(\mathrm{0}−\frac{\mathrm{17}}{\mathrm{7}}\right)^{\mathrm{2}} +\left(\mathrm{1}−\frac{\mathrm{13}}{\mathrm{7}}\right)^{\mathrm{2}} } \\ $$$$\mathrm{r}\:=\sqrt{\frac{\mathrm{289}+\mathrm{36}}{\mathrm{49}}}\:=\:\sqrt{\frac{\mathrm{325}}{\mathrm{49}}} \\ $$$$\mathrm{Therefore}\:\mathrm{the}\:\mathrm{circle}'\mathrm{s}\:\mathrm{equation}\:\mathrm{is}\: \\ $$$$\underline{\left(\mathrm{x}−\frac{\mathrm{17}}{\mathrm{7}}\right)^{\mathrm{2}} +\left(\mathrm{y}−\frac{\mathrm{13}}{\mathrm{7}}\right)^{\mathrm{2}} =\:\frac{\mathrm{325}}{\mathrm{49}}}\: \\ $$$$ \\ $$

Commented by bramlexs22 last updated on 07/Mar/21

Commented by bramlexs22 last updated on 07/Mar/21

$$\mathrm{nice}…. \\ $$

Commented by greg_ed last updated on 07/Mar/21

$$\boldsymbol{\mathrm{well}}\:\boldsymbol{\mathrm{done}},\:\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{EDWIN}}\mathrm{88}\:! \\ $$

Answered by john_santu last updated on 07/Mar/21

Let C(x,y) be a center point the circle we get the equation (i) (√(x^2 +(y−1)^2 )) = (√((x−1)^2 +(y−4)^2 )) (ii)(√(x^2 +(y−1)^2 )) =(√((x−5)^2 +(y−2)^2 )) (i)→x^2 +y^2 −2y+1=x^2 −2x+1+y^2 −8y+16 ⇒2x+6y=16 ; x+3y=8 (ii)→x^2 +y^2 −2y+1=x^2 −10x+25+y^2 −4y+4 ⇒10x+2y=28 ; 5x+y=14 solve for two equation ⇒x+3(14−5x)= 8 ⇒42−14x = 8 ; x=((34)/(14)) = ((17)/7) and y = 14−((85)/7) = ((13)/7) then we find radius =(√((((17)/7)−0)^2 +(((13)/7)−1)^2 )) radius =(√((289+36)/(49))) = (√((325)/(49))) ∴ (x−((17)/7))^2 +(y−((13)/7))^2 =((325)/(49))

$${Let}\:{C}\left({x},{y}\right)\:{be}\:{a}\:{center}\:{point}\:{the}\:{circle} \\ $$$${we}\:{get}\:{the}\:{equation}\: \\ $$$$\left({i}\right)\:\sqrt{{x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} }\:=\:\sqrt{\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{4}\right)^{\mathrm{2}} } \\ $$$$\left({ii}\right)\sqrt{{x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} }\:=\sqrt{\left({x}−\mathrm{5}\right)^{\mathrm{2}} +\left({y}−\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$\left({i}\right)\rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{y}+\mathrm{1}={x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}+{y}^{\mathrm{2}} −\mathrm{8}{y}+\mathrm{16} \\ $$$$\Rightarrow\mathrm{2}{x}+\mathrm{6}{y}=\mathrm{16}\:;\:{x}+\mathrm{3}{y}=\mathrm{8}\: \\ $$$$\left({ii}\right)\rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{y}+\mathrm{1}={x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{25}+{y}^{\mathrm{2}} −\mathrm{4}{y}+\mathrm{4} \\ $$$$\Rightarrow\mathrm{10}{x}+\mathrm{2}{y}=\mathrm{28}\:;\:\mathrm{5}{x}+{y}=\mathrm{14} \\ $$$${solve}\:{for}\:{two}\:{equation}\: \\ $$$$\Rightarrow{x}+\mathrm{3}\left(\mathrm{14}−\mathrm{5}{x}\right)=\:\mathrm{8} \\ $$$$\Rightarrow\mathrm{42}−\mathrm{14}{x}\:=\:\mathrm{8}\:;\:{x}=\frac{\mathrm{34}}{\mathrm{14}}\:=\:\frac{\mathrm{17}}{\mathrm{7}} \\ $$$${and}\:{y}\:=\:\mathrm{14}−\frac{\mathrm{85}}{\mathrm{7}}\:=\:\frac{\mathrm{13}}{\mathrm{7}} \\ $$$${then}\:{we}\:{find}\:{radius}\:=\sqrt{\left(\frac{\mathrm{17}}{\mathrm{7}}−\mathrm{0}\right)^{\mathrm{2}} +\left(\frac{\mathrm{13}}{\mathrm{7}}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${radius}\:=\sqrt{\frac{\mathrm{289}+\mathrm{36}}{\mathrm{49}}}\:=\:\sqrt{\frac{\mathrm{325}}{\mathrm{49}}} \\ $$$$\therefore\:\left({x}−\frac{\mathrm{17}}{\mathrm{7}}\right)^{\mathrm{2}} +\left({y}−\frac{\mathrm{13}}{\mathrm{7}}\right)^{\mathrm{2}} =\frac{\mathrm{325}}{\mathrm{49}} \\ $$

Answered by mr W last updated on 07/Mar/21

say the eqn. of circle is (x−h)^2 +(y−k)^2 =r^2 (0−h)^2 +(1−k)^2 =r^2 ...(i) (1−h)^2 +(4−k)^2 =r^2 ...(ii) (5−h)^2 +(2−k)^2 =r^2 ...(iii) (ii)−(i): (1−2h)(1)+(5−2k)(3)=0 ⇒h+3k=8 ...(I) (iii)−(i): (5−2h)(5)+(3−2k)(1)=0 ⇒5h+k=14 ...(II) from (I) and (II): ⇒k=((13)/7) ⇒h=((17)/7) from (i): ⇒r^2 =(((17)/7))^2 +(1−((13)/7))^2 =((325)/(49)) eqn. of circle: (x−((17)/7))^2 +(y−((13)/7))^2 =((325)/(49))

$${say}\:{the}\:{eqn}.\:{of}\:{circle}\:{is} \\ $$$$\left({x}−{h}\right)^{\mathrm{2}} +\left({y}−{k}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$ \\ $$$$\left(\mathrm{0}−{h}\right)^{\mathrm{2}} +\left(\mathrm{1}−{k}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \:\:\:…\left({i}\right) \\ $$$$\left(\mathrm{1}−{h}\right)^{\mathrm{2}} +\left(\mathrm{4}−{k}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \:\:\:…\left({ii}\right) \\ $$$$\left(\mathrm{5}−{h}\right)^{\mathrm{2}} +\left(\mathrm{2}−{k}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \:\:\:…\left({iii}\right) \\ $$$$\left({ii}\right)−\left({i}\right): \\ $$$$\left(\mathrm{1}−\mathrm{2}{h}\right)\left(\mathrm{1}\right)+\left(\mathrm{5}−\mathrm{2}{k}\right)\left(\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow{h}+\mathrm{3}{k}=\mathrm{8}\:\:\:…\left({I}\right) \\ $$$$\left({iii}\right)−\left({i}\right): \\ $$$$\left(\mathrm{5}−\mathrm{2}{h}\right)\left(\mathrm{5}\right)+\left(\mathrm{3}−\mathrm{2}{k}\right)\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{5}{h}+{k}=\mathrm{14}\:\:\:…\left({II}\right) \\ $$$${from}\:\left({I}\right)\:{and}\:\left({II}\right): \\ $$$$\Rightarrow{k}=\frac{\mathrm{13}}{\mathrm{7}} \\ $$$$\Rightarrow{h}=\frac{\mathrm{17}}{\mathrm{7}} \\ $$$${from}\:\left({i}\right): \\ $$$$\Rightarrow{r}^{\mathrm{2}} =\left(\frac{\mathrm{17}}{\mathrm{7}}\right)^{\mathrm{2}} +\left(\mathrm{1}−\frac{\mathrm{13}}{\mathrm{7}}\right)^{\mathrm{2}} =\frac{\mathrm{325}}{\mathrm{49}} \\ $$$${eqn}.\:{of}\:{circle}: \\ $$$$\left({x}−\frac{\mathrm{17}}{\mathrm{7}}\right)^{\mathrm{2}} +\left({y}−\frac{\mathrm{13}}{\mathrm{7}}\right)^{\mathrm{2}} =\frac{\mathrm{325}}{\mathrm{49}} \\ $$

Commented by mr W last updated on 07/Mar/21

Commented by bramlexs22 last updated on 07/Mar/21

how do you sir make a point in your graph?

$$\mathrm{how}\:\mathrm{do}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{make}\:\mathrm{a}\:\mathrm{point}\:\mathrm{in}\:\mathrm{your} \\ $$$$\mathrm{graph}? \\ $$

Commented by mr W last updated on 07/Mar/21

you just make an other smaller circle, for example with a radius of 0.01. the circle is (x−17/7)^2 +(y−13/7)^2 =325/49 the point is (x−17/7)^2 +(y−13/7)^2 <=0.01

$${you}\:{just}\:{make}\:{an}\:{other}\:{smaller}\:{circle},\: \\ $$$${for}\:{example}\:{with}\:{a}\:{radius}\:{of}\:\mathrm{0}.\mathrm{01}. \\ $$$${the}\:{circle}\:{is} \\ $$$$\left({x}−\mathrm{17}/\mathrm{7}\right)^{\mathrm{2}} +\left({y}−\mathrm{13}/\mathrm{7}\right)^{\mathrm{2}} =\mathrm{325}/\mathrm{49} \\ $$$${the}\:{point}\:{is} \\ $$$$\left({x}−\mathrm{17}/\mathrm{7}\right)^{\mathrm{2}} +\left({y}−\mathrm{13}/\mathrm{7}\right)^{\mathrm{2}} <=\mathrm{0}.\mathrm{01} \\ $$

Commented by mr W last updated on 07/Mar/21