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Find-the-equation-of-a-circle-which-touches-the-line-x-3y-13-0-and-passes-through-the-points-6-3-and-4-1-

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Question Number 170890 by nadovic last updated on 02/Jun/22
Find the equation of a circle which touches the line x−3y+13 = 0 and passes through the points (6, 3) and (4, −1).
Findtheequationofacirclewhichtouchesthelinex−3y+13=0andpassesthroughthepoints(6,3)and(4,−1).
Answered by aleks041103 last updated on 02/Jun/22
let the eqn be: c:(x−a)^2 +(y−b)^2 =r^2 then (6−a)^2 +(3−b)^2 =r^2 (4−a)^2 +(1+b)^2 =r^2 ⇒(6−a−4+a)(6−a+4−a)+(3−b−1−b)(3−b+1+b)=0 ⇒2(10−2a)+4(2−2b)=0 5−a+2−2b=0 ⇒a=7−2b ⇒(4−7+2b)^2 +(1+b)^2 =r^2 (2b−3)^2 +(1+b)^2 =r^2 4b^2 −12b+9+b^2 +2b+1=r^2 ⇒5b^2 −10b+10=r^2 ⇒c:(x+2b−7)^2 +(y−b)^2 =5b^2 −10b+10 x^2 +4b^2 +49+4bx−14x−28b+y^2 −2yb+b^2 =5b^2 −10b+10 x^2 +y^2 +(4b−14)x−2by+5b^2 −28b+49=5b^2 −10b+10 ⇒c:x^2 +y^2 +(4b−14)x−2by−18b+39=0 to touch line means for the system to have only one solution ⇒x−3y+13=0 ⇒x=3y−13 ⇒x^2 =9y^2 −78y+169 ⇒10y^2 −78y+169+(4b−14)(3y−13)−2by−18b+39=0 (4b−14)(3y−13)=12by−52b−42y+182 ⇒10y^2 +(−78+12b−42−2b)y+(169−52b+182−18b+39)=0 10y^2 +(10b−120)y+(390−70b)=0 y^2 +(b−12)y+(39−7b)=0 only one sol ⇒D=(b−12)^2 −4(39−7b)=0 ⇒b^2 −24b+144+28b−156=0 b^2 +4b−12=0 (b−2)(b+6)=0 ⇒b=−6,2 ⇒a=19,3 (a=7−2b) ⇒r^2 =5(36+12+2),5(4−4+2) (r^2 =5(b^2 −2b+2)) ⇒r=5(√(10)),(√(10)) ⇒c_1 :(x−19)^2 +(y+6)^2 =250 ⇒c_2 :(x−3)^2 +(y−2)^2 =10
lettheeqnbe:c:(x−a)2+(y−b)2=r2then(6−a)2+(3−b)2=r2(4−a)2+(1+b)2=r2⇒(6−a−4+a)(6−a+4−a)+(3−b−1−b)(3−b+1+b)=0⇒2(10−2a)+4(2−2b)=05−a+2−2b=0⇒a=7−2b⇒(4−7+2b)2+(1+b)2=r2(2b−3)2+(1+b)2=r24b2−12b+9+b2+2b+1=r2⇒5b2−10b+10=r2⇒c:(x+2b−7)2+(y−b)2=5b2−10b+10x2+4b2+49+4bx−14x−28b+y2−2yb+b2=5b2−10b+10x2+y2+(4b−14)x−2by+5b2−28b+49=5b2−10b+10⇒c:x2+y2+(4b−14)x−2by−18b+39=0totouchlinemeansforthesystemtohaveonlyonesolution⇒x−3y+13=0⇒x=3y−13⇒x2=9y2−78y+169⇒10y2−78y+169+(4b−14)(3y−13)−2by−18b+39=0(4b−14)(3y−13)=12by−52b−42y+182⇒10y2+(−78+12b−42−2b)y+(169−52b+182−18b+39)=010y2+(10b−120)y+(390−70b)=0y2+(b−12)y+(39−7b)=0onlyonesol⇒D=(b−12)2−4(39−7b)=0⇒b2−24b+144+28b−156=0b2+4b−12=0(b−2)(b+6)=0⇒b=−6,2⇒a=19,3(a=7−2b)⇒r2=5(36+12+2),5(4−4+2)(r2=5(b2−2b+2))⇒r=510,10⇒c1:(x−19)2+(y+6)2=250⇒c2:(x−3)2+(y−2)2=10
Commented by aleks041103 last updated on 02/Jun/22
Commented by nadovic last updated on 03/Jun/22
Thank you Sir
ThankyouSir
Answered by cortano1 last updated on 03/Jun/22
let (a,b) is the center point of circle (1)(√((a−6)^2 +(b−3)^2 )) =(√((a−4)^2 +(b+1)^2 )) ⇒−12a−6b+36+9=−8a+2b+16+1 ⇒−4a−8b=−28 ; a+2b=7; a=7−2b (2) ((∣a−3b+13∣)/( (√(10)))) =(√((a−4)^2 +(b+1)^2 )) ⇒((∣20−5b∣)/( (√(10)))) =(√((3−2b)^2 +(b+1)^2 )) ⇒∣20−5b∣=(√(10)) (√(5b^2 −10b+10)) ⇒∣4−b∣=(√(2b^2 −4b+4)) ⇒b^2 −8b+16=2b^2 −4b+4 ⇒b^2 +4b−12=0 ; { ((b=−6 ∧ a=19)),((b=2 ∧ a=3)) :} C_1 (19,−6) C_2 (3,2)
let(a,b)isthecenterpointofcircle(1)(a−6)2+(b−3)2=(a−4)2+(b+1)2⇒−12a−6b+36+9=−8a+2b+16+1⇒−4a−8b=−28;a+2b=7;a=7−2b(2)∣a−3b+13∣10=(a−4)2+(b+1)2⇒∣20−5b∣10=(3−2b)2+(b+1)2⇒∣20−5b∣=105b2−10b+10⇒∣4−b∣=2b2−4b+4⇒b2−8b+16=2b2−4b+4⇒b2+4b−12=0;{b=−6∧a=19b=2∧a=3C1(19,−6)C2(3,2)
Commented by nadovic last updated on 03/Jun/22
Thank you Sir
ThankyouSir

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