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Question-53689

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Question Number 53689 by ajfour last updated on 25/Jan/19
Commented by ajfour last updated on 25/Jan/19
Find R in terms of a,b, and r.
$${Find}\:{R}\:{in}\:{terms}\:{of}\:{a},{b},\:{and}\:{r}. \\ $$
Answered by mr W last updated on 25/Jan/19
A(0, 2r−b) C(0, −k) k=R+b−2r⇒R=k+2r−b eqn. of big circle: x^2 +(y+k)^2 =R^2 (x^2 /a^2 )+(y^2 /b^2 )=1 ((R^2 −(y+k)^2 )/a^2 )+(y^2 /b^2 )=1 b^2 R^2 −b^2 y^2 −b^2 k^2 −2kb^2 y+a^2 y^2 =a^2 b^2 (a^2 −b^2 )y^2 −2kb^2 y+b^2 (R^2 −a^2 −k^2 )=0 Δ=4k^2 b^4 −4(a^2 −b^2 )b^2 (R^2 −a^2 −k^2 )=0 k^2 b^2 −(a^2 −b^2 )(R^2 −a^2 −k^2 )=0 k^2 b^2 −(a^2 −b^2 )[(k+2r−b)^2 −a^2 −k^2 ]=0 ⇒b^2 k^2 −2(a^2 −b^2 )(2r−b)k−(a^2 −b^2 )(2r−b−a)(2r−b+a)=0 ⇒k=(((a^2 −b^2 )(2r−b)±(√((a^2 −b^2 )[(a^2 −b^2 )(2r−b)^2 +b^2 (2r−b−a)(2r−b+a)])))/b^2 ) ⇒k=(((a^2 −b^2 )(2r−b)±2a(√((a^2 −b^2 )(r−b)r)))/b^2 ) ⇒R=2r−b+(((a^2 −b^2 )(2r−b)±2a(√((a^2 −b^2 )(r−b)r)))/b^2 ) ⇒R=((a^2 (2r−b)±2a(√((a^2 −b^2 )(r−b)r)))/b^2 )
$${A}\left(\mathrm{0},\:\mathrm{2}{r}−{b}\right) \\ $$$${C}\left(\mathrm{0},\:−{k}\right) \\ $$$${k}={R}+{b}−\mathrm{2}{r}\Rightarrow{R}={k}+\mathrm{2}{r}−{b} \\ $$$$ \\ $$$${eqn}.\:{of}\:{big}\:{circle}: \\ $$$${x}^{\mathrm{2}} +\left({y}+{k}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{R}^{\mathrm{2}} −\left({y}+{k}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${b}^{\mathrm{2}} {R}^{\mathrm{2}} −{b}^{\mathrm{2}} {y}^{\mathrm{2}} −{b}^{\mathrm{2}} {k}^{\mathrm{2}} −\mathrm{2}{kb}^{\mathrm{2}} {y}+{a}^{\mathrm{2}} {y}^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){y}^{\mathrm{2}} −\mathrm{2}{kb}^{\mathrm{2}} {y}+{b}^{\mathrm{2}} \left({R}^{\mathrm{2}} −{a}^{\mathrm{2}} −{k}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Delta=\mathrm{4}{k}^{\mathrm{2}} {b}^{\mathrm{4}} −\mathrm{4}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){b}^{\mathrm{2}} \left({R}^{\mathrm{2}} −{a}^{\mathrm{2}} −{k}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${k}^{\mathrm{2}} {b}^{\mathrm{2}} −\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left({R}^{\mathrm{2}} −{a}^{\mathrm{2}} −{k}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${k}^{\mathrm{2}} {b}^{\mathrm{2}} −\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left[\left({k}+\mathrm{2}{r}−{b}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} −{k}^{\mathrm{2}} \right]=\mathrm{0} \\ $$$$\Rightarrow{b}^{\mathrm{2}} {k}^{\mathrm{2}} −\mathrm{2}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left(\mathrm{2}{r}−{b}\right){k}−\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left(\mathrm{2}{r}−{b}−{a}\right)\left(\mathrm{2}{r}−{b}+{a}\right)=\mathrm{0} \\ $$$$\Rightarrow{k}=\frac{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left(\mathrm{2}{r}−{b}\right)\pm\sqrt{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left[\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left(\mathrm{2}{r}−{b}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \left(\mathrm{2}{r}−{b}−{a}\right)\left(\mathrm{2}{r}−{b}+{a}\right)\right]}}{{b}^{\mathrm{2}} } \\ $$$$\Rightarrow{k}=\frac{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left(\mathrm{2}{r}−{b}\right)\pm\mathrm{2}{a}\sqrt{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left({r}−{b}\right){r}}}{{b}^{\mathrm{2}} } \\ $$$$\Rightarrow{R}=\mathrm{2}{r}−{b}+\frac{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left(\mathrm{2}{r}−{b}\right)\pm\mathrm{2}{a}\sqrt{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left({r}−{b}\right){r}}}{{b}^{\mathrm{2}} } \\ $$$$\Rightarrow{R}=\frac{{a}^{\mathrm{2}} \left(\mathrm{2}{r}−{b}\right)\pm\mathrm{2}{a}\sqrt{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left({r}−{b}\right){r}}}{{b}^{\mathrm{2}} } \\ $$
Commented by ajfour last updated on 25/Jan/19
thanks, Sir. How to choose among the signs?
$${thanks},\:{Sir}.\:{How}\:{to}\:{choose}\:{among}\:{the}\:{signs}? \\ $$
Commented by mr W last updated on 25/Jan/19
take + − is for the small circle under point A btw. small circle and ellipse.
$${take}\:+ \\ $$$$−\:{is}\:{for}\:{the}\:{small}\:{circle}\:{under}\:{point}\:{A} \\ $$$${btw}.\:{small}\:{circle}\:{and}\:{ellipse}. \\ $$
Commented by ajfour last updated on 25/Jan/19
Yes sir, understood, as i attempted.
$${Yes}\:{sir},\:{understood},\:{as}\:{i}\:{attempted}. \\ $$
Answered by ajfour last updated on 25/Jan/19
eq. of ellipse x^2 +(a^2 /b^2 )y^2 =a^2 C(0,−k) 2r−(b−k)=R ⇒ k=R−2r+b ...(i) eq. of circumcircle x^2 +(y+k)^2 =R^2 As ellipse and circle graze, ⇒ R^2 −(y+k)^2 +((a^2 y^2 )/b^2 )=a^2 should yield double root for y. y^2 ((a^2 /b^2 )−1)−2ky+R^2 −a^2 −k^2 = 0 D=0, ⇒ 4k^2 =4((a^2 /b^2 )−1)(R^2 −a^2 −k^2 ) ⇒ ((a^2 k^2 )/b^2 ) = (((R^2 −a^2 )(a^2 −b^2 ))/b^2 ) ⇒ a^2 (R−2r+b)^2 =(a^2 −b^2 )(R^2 −a^2 ) ⇒ b^2 R^2 −2a^2 (2r−b)R +a^2 (2r−b)^2 +a^4 −a^2 b^2 = 0 ⇒ b^2 R^2 −2a^2 (2r−b)R+4a^2 r(r−b)=0 ⇒ R = ((a^2 (2r−b)±a(√(a^2 (2r−b)^2 −4b^2 r(r−b))))/b^2 ) R= (a^2 /b^2 ){(2r−b)±(√((2r−b)^2 −((4b^2 )/a^2 )r(r−b))) } Its clear we′ll choose +ve sign for the circumcircle in diagram. The −ve one is for the circle that′s tangent to ellipse from within the small circle(tangent still to it at A).
$${eq}.\:{of}\:{ellipse}\:\:{x}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$${C}\left(\mathrm{0},−{k}\right) \\ $$$$\mathrm{2}{r}−\left({b}−{k}\right)={R} \\ $$$$\Rightarrow\:{k}={R}−\mathrm{2}{r}+{b}\:\:\:…\left({i}\right) \\ $$$${eq}.\:{of}\:{circumcircle} \\ $$$$\:\:\:\:{x}^{\mathrm{2}} +\left({y}+{k}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${As}\:{ellipse}\:{and}\:{circle}\:{graze}, \\ $$$$\Rightarrow\:{R}^{\mathrm{2}} −\left({y}+{k}\right)^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} {y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }={a}^{\mathrm{2}} \:\:\:{should} \\ $$$${yield}\:{double}\:{root}\:{for}\:{y}. \\ $$$${y}^{\mathrm{2}} \left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−\mathrm{1}\right)−\mathrm{2}{ky}+{R}^{\mathrm{2}} −{a}^{\mathrm{2}} −{k}^{\mathrm{2}} =\:\mathrm{0} \\ $$$${D}=\mathrm{0},\:\:\:\Rightarrow \\ $$$$\:\:\:\:\mathrm{4}{k}^{\mathrm{2}} =\mathrm{4}\left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−\mathrm{1}\right)\left({R}^{\mathrm{2}} −{a}^{\mathrm{2}} −{k}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:\frac{{a}^{\mathrm{2}} {k}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=\:\frac{\left({R}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}{{b}^{\mathrm{2}} } \\ $$$$\Rightarrow\:{a}^{\mathrm{2}} \left({R}−\mathrm{2}{r}+{b}\right)^{\mathrm{2}} =\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left({R}^{\mathrm{2}} −{a}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:{b}^{\mathrm{2}} {R}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} \left(\mathrm{2}{r}−{b}\right){R} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{a}^{\mathrm{2}} \left(\mathrm{2}{r}−{b}\right)^{\mathrm{2}} +{a}^{\mathrm{4}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\Rightarrow\:{b}^{\mathrm{2}} {R}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} \left(\mathrm{2}{r}−{b}\right){R}+\mathrm{4}{a}^{\mathrm{2}} {r}\left({r}−{b}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{R}\:=\:\frac{{a}^{\mathrm{2}} \left(\mathrm{2}{r}−{b}\right)\pm{a}\sqrt{{a}^{\mathrm{2}} \left(\mathrm{2}{r}−{b}\right)^{\mathrm{2}} −\mathrm{4}{b}^{\mathrm{2}} {r}\left({r}−{b}\right)}}{{b}^{\mathrm{2}} } \\ $$$${R}=\:\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\left\{\left(\mathrm{2}{r}−{b}\right)\pm\sqrt{\left(\mathrm{2}{r}−{b}\right)^{\mathrm{2}} −\frac{\mathrm{4}{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }{r}\left({r}−{b}\right)}\:\right\} \\ $$$${Its}\:{clear}\:{we}'{ll}\:{choose}\:+{ve}\:{sign} \\ $$$${for}\:{the}\:{circumcircle}\:{in}\:{diagram}. \\ $$$${The}\:−{ve}\:{one}\:{is}\:{for}\:{the}\:{circle}\:{that}'{s} \\ $$$${tangent}\:{to}\:{ellipse}\:{from}\:{within}\:{the} \\ $$$${small}\:{circle}\left({tangent}\:{still}\:{to}\:{it}\:{at}\:{A}\right). \\ $$
Commented by mr W last updated on 25/Jan/19
very fine sir!
$${very}\:{fine}\:{sir}! \\ $$

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