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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
$$\left(\mathrm{1}\right)\:\mathrm{let}\:\left(\mathrm{a},\mathrm{b}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{center}\:\mathrm{point}\:\mathrm{of} \\ $$$$\mathrm{circle} \\ $$$$\left(\mathrm{2}\right)\:\left(\frac{\mathrm{b}−\mathrm{1}}{\mathrm{a}+\mathrm{1}}\right).\left(−\mathrm{3}\right)=−\mathrm{1} \\ $$$$\Rightarrow\mathrm{3b}−\mathrm{3}=\mathrm{a}+\mathrm{1}\:;\:\mathrm{a}=\mathrm{3b}−\mathrm{4} \\ $$$$\left(\mathrm{3}\right)\:\frac{\mid\mathrm{3a}+\mathrm{b}+\mathrm{2}\mid}{\:\sqrt{\mathrm{10}}}\:=\:\sqrt{\left(\mathrm{a}−\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{b}−\mathrm{5}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\mid\mathrm{10b}−\mathrm{10}\mid=\sqrt{\mathrm{10}}\:\sqrt{\left(\mathrm{3b}−\mathrm{7}\right)^{\mathrm{2}} +\left(\mathrm{b}−\mathrm{5}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:\mathrm{10}\left(\mathrm{b}^{\mathrm{2}} −\mathrm{2b}+\mathrm{1}\right)=\mathrm{10b}^{\mathrm{2}} −\mathrm{10b}−\mathrm{42b}+\mathrm{74} \\ $$$$\Rightarrow−\mathrm{20b}+\mathrm{10}=−\mathrm{52b}+\mathrm{74} \\ $$$$\Rightarrow\mathrm{32b}\:=\:\mathrm{64}\:\rightarrow\begin{cases}{\mathrm{b}=\mathrm{2}}\\{\mathrm{a}=\mathrm{2}}\end{cases} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{radius}\:=\:\sqrt{\left(\mathrm{2}−\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{2}−\mathrm{5}\right)^{\mathrm{2}} }\:=\sqrt{\mathrm{10}} \\ $$$$\left(\mathrm{5}\right)\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{circle} \\ $$$$\therefore\:\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{10} \\ $$

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