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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)?\\n

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

$$\left(\mathrm{1}\right)\:\mathrm{let}\:\left(\mathrm{a},\mathrm{b}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{center}\:\mathrm{point}\:\mathrm{of} \\ $$ $$\mathrm{circle} \\ $$ $$\left(\mathrm{2}\right)\:\left(\frac{\mathrm{b}−\mathrm{1}}{\mathrm{a}+\mathrm{1}}\right).\left(−\mathrm{3}\right)=−\mathrm{1} \\ $$ $$\Rightarrow\mathrm{3b}−\mathrm{3}=\mathrm{a}+\mathrm{1}\:;\:\mathrm{a}=\mathrm{3b}−\mathrm{4} \\ $$ $$\left(\mathrm{3}\right)\:\frac{\mid\mathrm{3a}+\mathrm{b}+\mathrm{2}\mid}{\:\sqrt{\mathrm{10}}}\:=\:\sqrt{\left(\mathrm{a}−\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{b}−\mathrm{5}\right)^{\mathrm{2}} } \\ $$ $$\Rightarrow\mid\mathrm{10b}−\mathrm{10}\mid=\sqrt{\mathrm{10}}\:\sqrt{\left(\mathrm{3b}−\mathrm{7}\right)^{\mathrm{2}} +\left(\mathrm{b}−\mathrm{5}\right)^{\mathrm{2}} } \\ $$ $$\Rightarrow\:\mathrm{10}\left(\mathrm{b}^{\mathrm{2}} −\mathrm{2b}+\mathrm{1}\right)=\mathrm{10b}^{\mathrm{2}} −\mathrm{10b}−\mathrm{42b}+\mathrm{74} \\ $$ $$\Rightarrow−\mathrm{20b}+\mathrm{10}=−\mathrm{52b}+\mathrm{74} \\ $$ $$\Rightarrow\mathrm{32b}\:=\:\mathrm{64}\:\rightarrow\begin{cases}{\mathrm{b}=\mathrm{2}}\\{\mathrm{a}=\mathrm{2}}\end{cases} \\ $$ $$\left(\mathrm{4}\right)\:\mathrm{radius}\:=\:\sqrt{\left(\mathrm{2}−\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{2}−\mathrm{5}\right)^{\mathrm{2}} }\:=\sqrt{\mathrm{10}} \\ $$ $$\left(\mathrm{5}\right)\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{circle} \\ $$ $$\therefore\:\color{mathblue}{\left(}\mathrm{\color{mathblue}{x}}\color{mathblue}{−}\mathrm{\color{mathblue}{2}}\color{mathblue}{\right)}^{\mathrm{\color{mathblue}{2}}} \color{mathblue}{+}\color{mathblue}{\left(}\mathrm{\color{mathblue}{y}}\color{mathblue}{−}\mathrm{\color{mathblue}{2}}\color{mathblue}{\right)}^{\mathrm{\color{mathblue}{2}}} \color{mathblue}{=}\mathrm{\color{mathblue}{1}\color{mathblue}{0}} \\ $$

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