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let-the-cercle-x-1-2-y-3-2-9-and-the-point-A-4-1-vrrify-that-A-is-out-of-circle-and-determine-the-equation-of-two-tangentes-to-circle-wich-passes-by-point-A-

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Question Number 77367 by msup trace by abdo last updated on 05/Jan/20
let the cercle (x+1)^(2 ) +(y−3)^2 =9 and the point A(4,1) vrrify that A is out of circle and determine the equation of two tangentes to circle wich passes by point A.
$${let}\:{the}\:{cercle}\:\:\left({x}+\mathrm{1}\right)^{\mathrm{2}\:} +\left({y}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{9} \\ $$$${and}\:{the}\:{point}\:\:{A}\left(\mathrm{4},\mathrm{1}\right) \\ $$$${vrrify}\:{that}\:\:{A}\:\:{is}\:{out}\:{of}\:{circle} \\ $$$${and}\:\:{determine}\:{the}\:{equation}\:{of} \\ $$$${two}\:{tangentes}\:{to}\:{circle}\:{wich} \\ $$$${passes}\:{by}\:{point}\:{A}. \\ $$
Commented by mathmax by abdo last updated on 05/Jan/20
the plan is provided with orthonormal reference(o,i^→ ,j^(→)) ).
$${the}\:{plan}\:{is}\:{provided}\:{with}\:{orthonormal}\:{reference}\left({o},\overset{\rightarrow} {{i}},\overset{\left.\rightarrow\right)} {{j}}\right). \\ $$
Answered by jagoll last updated on 06/Jan/20
test point A(4,1) ⇒ 5^2 +(−2)^2 >9 then A is out the circle (2) let y = mx + n is tangent the circle substitute point A ⇒ 1=4m+n n = 1−4m. rewrite tangent line y = mx + 1−4m distance of center circle to line equal to radius ⇒ 3 =∣((−m+1−4m−3)/( (√(1+m^2 ))))∣ 3(√(1+m^2 )) = ∣−5m−2∣ 9+9m^2 =25m^2 +20m+4 16m^2 +20m−5=0 m_1 = (1/4) and m_2 = −(5/4). now you get the tangent line.
$$\mathrm{test}\:\mathrm{point}\:\mathrm{A}\left(\mathrm{4},\mathrm{1}\right)\:\Rightarrow\:\mathrm{5}^{\mathrm{2}} +\left(−\mathrm{2}\right)^{\mathrm{2}} >\mathrm{9} \\ $$$$\mathrm{then}\:\mathrm{A}\:\mathrm{is}\:\mathrm{out}\:\mathrm{the}\:\mathrm{circle} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{let}\:\mathrm{y}\:=\:\mathrm{mx}\:+\:\mathrm{n}\:\mathrm{is}\:\mathrm{tangent}\:\mathrm{the}\:\mathrm{circle} \\ $$$$\mathrm{substitute}\:\mathrm{point}\:\mathrm{A}\:\Rightarrow\:\mathrm{1}=\mathrm{4m}+\mathrm{n} \\ $$$$\mathrm{n}\:=\:\mathrm{1}−\mathrm{4m}.\:\mathrm{rewrite}\:\mathrm{tangent}\:\mathrm{line}\: \\ $$$$\mathrm{y}\:=\:\mathrm{mx}\:+\:\mathrm{1}−\mathrm{4m}\: \\ $$$$\mathrm{distance}\:\mathrm{of}\:\mathrm{center}\:\mathrm{circle}\:\mathrm{to}\:\mathrm{line}\:\mathrm{equal} \\ $$$$\mathrm{to}\:\mathrm{radius}\:\Rightarrow\:\mathrm{3}\:=\mid\frac{−\mathrm{m}+\mathrm{1}−\mathrm{4m}−\mathrm{3}}{\:\sqrt{\mathrm{1}+\mathrm{m}^{\mathrm{2}} }}\mid \\ $$$$\mathrm{3}\sqrt{\mathrm{1}+\mathrm{m}^{\mathrm{2}} }\:=\:\mid−\mathrm{5m}−\mathrm{2}\mid \\ $$$$\mathrm{9}+\mathrm{9m}^{\mathrm{2}} =\mathrm{25m}^{\mathrm{2}} +\mathrm{20m}+\mathrm{4} \\ $$$$\mathrm{16m}^{\mathrm{2}} +\mathrm{20m}−\mathrm{5}=\mathrm{0} \\ $$$$\mathrm{m}_{\mathrm{1}} =\:\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{and}\:\mathrm{m}_{\mathrm{2}} \:=\:−\frac{\mathrm{5}}{\mathrm{4}}.\:\mathrm{now}\:\mathrm{you} \\ $$$$\mathrm{get}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{line}.\: \\ $$
Commented by msup trace by abdo last updated on 06/Jan/20
thanks sir.
$${thanks}\:{sir}. \\ $$

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