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Find-the-equation-of-the-line-which-passes-through-the-point-3-5-and-is-tangent-to-the-circle-x-1-2-y-1-2-4-

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Question Number 183044 by depressiveshrek last updated on 19/Dec/22
Find the equation of the line which passes through the point (3, 5) and is tangent to the circle (x−1)^2 +(y−1)^2 =4
$${Find}\:{the}\:{equation}\:{of}\:{the}\:{line}\:{which} \\ $$$${passes}\:{through}\:{the}\:{point}\:\left(\mathrm{3},\:\mathrm{5}\right) \\ $$$${and}\:{is}\:{tangent}\:{to}\:{the}\:{circle} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4} \\ $$
Answered by mr W last updated on 19/Dec/22
say line through (3,5) is: x−3=k(y−5) ⇒x−ky+5k−3=0 ((∣1−k+5k−3∣)/( (√(1+k^2 ))))=2 ((∣2k−1∣)/( (√(1+k^2 ))))=1 (3k−4)k=0 ⇒k=0 or k=(4/3) line 1: x−3=0 ⇒x=3 line 2: x−3=(4/3)(y−5) ⇒y=((3x)/4)+((11)/4)
$${say}\:{line}\:{through}\:\left(\mathrm{3},\mathrm{5}\right)\:{is}:\: \\ $$$${x}−\mathrm{3}={k}\left({y}−\mathrm{5}\right) \\ $$$$\Rightarrow{x}−{ky}+\mathrm{5}{k}−\mathrm{3}=\mathrm{0} \\ $$$$\frac{\mid\mathrm{1}−{k}+\mathrm{5}{k}−\mathrm{3}\mid}{\:\sqrt{\mathrm{1}+{k}^{\mathrm{2}} }}=\mathrm{2} \\ $$$$\frac{\mid\mathrm{2}{k}−\mathrm{1}\mid}{\:\sqrt{\mathrm{1}+{k}^{\mathrm{2}} }}=\mathrm{1} \\ $$$$\left(\mathrm{3}{k}−\mathrm{4}\right){k}=\mathrm{0} \\ $$$$\Rightarrow{k}=\mathrm{0}\:{or}\:{k}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${line}\:\mathrm{1}:\:{x}−\mathrm{3}=\mathrm{0}\:\Rightarrow{x}=\mathrm{3} \\ $$$${line}\:\mathrm{2}:\:{x}−\mathrm{3}=\frac{\mathrm{4}}{\mathrm{3}}\left({y}−\mathrm{5}\right)\:\Rightarrow{y}=\frac{\mathrm{3}{x}}{\mathrm{4}}+\frac{\mathrm{11}}{\mathrm{4}} \\ $$
Commented by mr W last updated on 19/Dec/22
Answered by MathAcer26 last updated on 19/Dec/22
The center is at (1,1) with radius = 2 (3,1) is a point in the circle. Thus, x = 3 is a tangent to the circle passing through (3,5)
$$\mathrm{The}\:\mathrm{center}\:\mathrm{is}\:\mathrm{at}\:\left(\mathrm{1},\mathrm{1}\right)\:\mathrm{with}\:\mathrm{radius}\:=\:\mathrm{2} \\ $$$$\left(\mathrm{3},\mathrm{1}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{point}\:\mathrm{in}\:\mathrm{the}\:\mathrm{circle}. \\ $$$$\mathrm{Thus},\:\mathrm{x}\:=\:\mathrm{3}\:\mathrm{is}\:\mathrm{a}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{passing} \\ $$$$\mathrm{through}\:\left(\mathrm{3},\mathrm{5}\right) \\ $$
Answered by ajfour last updated on 20/Dec/22
let Origin (1,1) circle x^2 +y^2 =4 tangent hx+ky=4 passes through (2,4) 2h+4k=4 ⇒ h=2−2k h^2 +k^2 =4 ⇒ 4(1−k)^2 +k^2 =4 5k^2 −8k=0 ⇒ k=(8/5) , h=−(6/5) eq.of tangent in O(0,0) h(x−1)+k(y−1)=4 −6x+8y=22 4y=3x+11 is the tangent equation with O(0,0).
$${let}\:{Origin}\:\left(\mathrm{1},\mathrm{1}\right) \\ $$$${circle}\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4} \\ $$$${tangent}\:{hx}+{ky}=\mathrm{4} \\ $$$${passes}\:{through}\:\left(\mathrm{2},\mathrm{4}\right) \\ $$$$\mathrm{2}{h}+\mathrm{4}{k}=\mathrm{4}\:\Rightarrow\:\:{h}=\mathrm{2}−\mathrm{2}{k} \\ $$$${h}^{\mathrm{2}} +{k}^{\mathrm{2}} =\mathrm{4} \\ $$$$\Rightarrow\:\mathrm{4}\left(\mathrm{1}−{k}\right)^{\mathrm{2}} +{k}^{\mathrm{2}} =\mathrm{4} \\ $$$$\mathrm{5}{k}^{\mathrm{2}} −\mathrm{8}{k}=\mathrm{0}\:\:\:\Rightarrow\:\:{k}=\frac{\mathrm{8}}{\mathrm{5}}\:\:,\:{h}=−\frac{\mathrm{6}}{\mathrm{5}} \\ $$$${eq}.{of}\:{tangent}\:{in}\:{O}\left(\mathrm{0},\mathrm{0}\right) \\ $$$${h}\left({x}−\mathrm{1}\right)+{k}\left({y}−\mathrm{1}\right)=\mathrm{4} \\ $$$$−\mathrm{6}{x}+\mathrm{8}{y}=\mathrm{22} \\ $$$$\mathrm{4}{y}=\mathrm{3}{x}+\mathrm{11} \\ $$$${is}\:{the}\:{tangent}\:{equation}\:{with} \\ $$$${O}\left(\mathrm{0},\mathrm{0}\right). \\ $$
Answered by cortano1 last updated on 20/Dec/22
let the line is (a−1)(x−1)+(b−1)(y−1)= 4 and passes throught the point (3,5) ⇒2(a−1)+4(b−1)= 4 ⇒a−1+2b−2=2 ; a=5−2b (ii) (a−1)^2 +(b−1)^2 =4 ⇒(4−2b)^2 +(b−1)^2 =4 ⇒ { ((b=1⇒a=3)),((b=((13)/5)⇒a=5−((26)/5)=−(1/5))) :} so the equation of line is ⇒ { ((2(x−1)=4⇒x=3)),((−(6/5)(x−1)+(8/5)(y−1)=4)) :} ⇒ { ((x=3)),((−6x+6+8y−8=20)) :} ⇒ { ((x=3)),((3x−4y+11=0)) :}
$$\:{let}\:{the}\:{line}\:{is}\:\left({a}−\mathrm{1}\right)\left({x}−\mathrm{1}\right)+\left({b}−\mathrm{1}\right)\left({y}−\mathrm{1}\right)=\:\mathrm{4} \\ $$$${and}\:{passes}\:{throught}\:{the}\:{point}\:\left(\mathrm{3},\mathrm{5}\right) \\ $$$$\Rightarrow\mathrm{2}\left({a}−\mathrm{1}\right)+\mathrm{4}\left({b}−\mathrm{1}\right)=\:\mathrm{4} \\ $$$$\Rightarrow{a}−\mathrm{1}+\mathrm{2}{b}−\mathrm{2}=\mathrm{2}\:;\:{a}=\mathrm{5}−\mathrm{2}{b} \\ $$$$\left({ii}\right)\:\left({a}−\mathrm{1}\right)^{\mathrm{2}} +\left({b}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4} \\ $$$$\Rightarrow\left(\mathrm{4}−\mathrm{2}{b}\right)^{\mathrm{2}} +\left({b}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4} \\ $$$$\Rightarrow\begin{cases}{{b}=\mathrm{1}\Rightarrow{a}=\mathrm{3}}\\{{b}=\frac{\mathrm{13}}{\mathrm{5}}\Rightarrow{a}=\mathrm{5}−\frac{\mathrm{26}}{\mathrm{5}}=−\frac{\mathrm{1}}{\mathrm{5}}}\end{cases} \\ $$$${so}\:{the}\:{equation}\:{of}\:{line}\:{is} \\ $$$$\Rightarrow\begin{cases}{\mathrm{2}\left({x}−\mathrm{1}\right)=\mathrm{4}\Rightarrow{x}=\mathrm{3}}\\{−\frac{\mathrm{6}}{\mathrm{5}}\left({x}−\mathrm{1}\right)+\frac{\mathrm{8}}{\mathrm{5}}\left({y}−\mathrm{1}\right)=\mathrm{4}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{{x}=\mathrm{3}}\\{−\mathrm{6}{x}+\mathrm{6}+\mathrm{8}{y}−\mathrm{8}=\mathrm{20}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{{x}=\mathrm{3}}\\{\mathrm{3}{x}−\mathrm{4}{y}+\mathrm{11}=\mathrm{0}}\end{cases} \\ $$

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