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What-is-the-equation-of-the-circle-if-the-circle-is-tangential-to-the-line-3x-y-2-0-at-1-1-and-it-passes-through-the-point-3-5-

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Question Number 140325 by liberty last updated on 06/May/21
What is the equation of the circle, if the circle is tangential to the line 3x+y+2=0 at (-1,1) and it passes through the point (3,5)?
What is the equation of the circle, if the circle is tangential to the line 3x+y+2=0 at (-1,1) and it passes through the point (3,5)?
Answered by benjo_mathlover last updated on 06/May/21
(1) let (a,b) is the center point of circle (2) (((b−1)/(a+1))).(−3)=−1 ⇒3b−3=a+1 ; a=3b−4 (3) ((∣3a+b+2∣)/( (√(10)))) = (√((a−3)^2 +(b−5)^2 )) ⇒∣10b−10∣=(√(10)) (√((3b−7)^2 +(b−5)^2 )) ⇒ 10(b^2 −2b+1)=10b^2 −10b−42b+74 ⇒−20b+10=−52b+74 ⇒32b = 64 → { ((b=2)),((a=2)) :} (4) radius = (√((2−3)^2 +(2−5)^2 )) =(√(10)) (5) the equation of circle ∴ (x−2)^2 +(y−2)^2 =10
(1)let(a,b)isthecenterpointofcircle(2)(b−1a+1).(−3)=−1⇒3b−3=a+1;a=3b−4(3)∣3a+b+2∣10=(a−3)2+(b−5)2⇒∣10b−10∣=10(3b−7)2+(b−5)2⇒10(b2−2b+1)=10b2−10b−42b+74⇒−20b+10=−52b+74⇒32b=64→{b=2a=2(4)radius=(2−3)2+(2−5)2=10(5)theequationofcircle∴(x−2)2+(y−2)2=10

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