Question Number 28512 by ajfour last updated on 26/Jan/18
Commented by ajfour last updated on 26/Jan/18
$${If}\:{eq}.\:{of}\:{line}\:{AB}\:{is}\:\boldsymbol{{y}}=\boldsymbol{{mx}}+\boldsymbol{{c}} \\ $$$${and}\:{that}\:{of}\:{ellipse}\:{is}\:\:\frac{\boldsymbol{{x}}^{\mathrm{2}} }{\boldsymbol{{a}}^{\mathrm{2}} }+\frac{\boldsymbol{{y}}^{\mathrm{2}} }{\boldsymbol{{b}}^{\mathrm{2}} }=\mathrm{1}\:, \\ $$$${find}\:{eq}.\:{of}\:{circle}\:{with}\:{AB}\:{as} \\ $$$${diameter}. \\ $$
Answered by mrW2 last updated on 26/Jan/18
![y=mx+c (x^2 /a^2 )+(y^2 /b^2 )=1 (x^2 /a^2 )+(((mx+c)^2 )/b^2 )=1 (a^2 m^2 +b^2 )x^2 +2a^2 mcx+a^2 (c^2 −b^2 )=0 ⇒x=((−a^2 mc±(√(a^4 m^2 c^2 −(a^2 m^2 +b^2 )a^2 (c^2 −b^2 ))))/(a^2 m^2 +b^2 )) ⇒x_(A,B) =((−a^2 mc±ab(√(a^2 m^2 +b^2 −c^2 )))/(a^2 m^2 +b^2 )) ⇒x_C =((x_A +x_B )/2)=−((a^2 mc)/(a^2 m^2 +b^2 )) ⇒y_C =((b^2 c)/(a^2 m^2 +b^2 )) 2R=∣x_A −x_B ∣(√(1+m^2 ))=((2ab(√((a^2 m^2 +b^2 −c^2 )(1+m^2 ))))/(a^2 m^2 +b^2 )) ⇒R=((ab(√((a^2 m^2 +b^2 −c^2 )(1+m^2 ))))/(a^2 m^2 +b^2 )) Eqn. of circle: (x−x_C )^2 +(y−y_C )^2 =R^2 ⇒(x+((a^2 mc)/(a^2 m^2 +b^2 )))^2 +(y−((b^2 c)/(a^2 m^2 +b^2 )))^2 =((a^2 b^2 (a^2 m^2 +b^2 −c^2 )(1+m^2 ))/((a^2 m^2 +b^2 )^2 )) ⇒[(a^2 m^2 +b^2 )x+a^2 mc]^2 +[(a^2 m^2 +b^2 )y−b^2 c]^2 =a^2 b^2 (a^2 m^2 +b^2 −c^2 )(1+m^2 ) ⇒(a^2 m^2 +b^2 )^2 (x^2 +y^2 )+2(a^2 m^2 +b^2 )a^2 mcx+a^4 m^2 c^2 −2(a^2 m^2 +b^2 )b^2 cy+b^4 c^2 =a^2 b^2 (a^2 m^2 +b^2 )(1+m^2 )−a^2 b^2 c^2 −a^2 b^2 c^2 m^2 ⇒(a^2 m^2 +b^2 )^2 (x^2 +y^2 )+2(a^2 m^2 +b^2 )(a^2 mcx−b^2 cy)+(a^2 m^2 +b^2 )a^2 c^2 −a^2 b^2 (a^2 m^2 +b^2 )(1+m^2 )+(a^2 m^2 +b^2 )b^2 c^2 =0 ⇒(a^2 m^2 +b^2 )(x^2 +y^2 )+2a^2 mcx−2b^2 cy+(a^2 +b^2 )c^2 −a^2 b^2 (1+m^2 )=0](https://www.tinkutara.com/question/Q28514.png)
$${y}={mx}+{c} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left({mx}+{c}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\left({a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){x}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{2}} {mcx}+{a}^{\mathrm{2}} \left({c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{−{a}^{\mathrm{2}} {mc}\pm\sqrt{{a}^{\mathrm{4}} {m}^{\mathrm{2}} {c}^{\mathrm{2}} −\left({a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){a}^{\mathrm{2}} \left({c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}}{{a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\Rightarrow{x}_{{A},{B}} =\frac{−{a}^{\mathrm{2}} {mc}\pm{ab}\sqrt{{a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }}{{a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\Rightarrow{x}_{{C}} =\frac{{x}_{{A}} +{x}_{{B}} }{\mathrm{2}}=−\frac{{a}^{\mathrm{2}} {mc}}{{a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\Rightarrow{y}_{{C}} =\frac{{b}^{\mathrm{2}} {c}}{{a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$ \\ $$$$\mathrm{2}{R}=\mid{x}_{{A}} −{x}_{{B}} \mid\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }=\frac{\mathrm{2}{ab}\sqrt{\left({a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)\left(\mathrm{1}+{m}^{\mathrm{2}} \right)}}{{a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\Rightarrow{R}=\frac{{ab}\sqrt{\left({a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)\left(\mathrm{1}+{m}^{\mathrm{2}} \right)}}{{a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$ \\ $$$${Eqn}.\:{of}\:{circle}: \\ $$$$\left({x}−{x}_{{C}} \right)^{\mathrm{2}} +\left({y}−{y}_{{C}} \right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\Rightarrow\left({x}+\frac{{a}^{\mathrm{2}} {mc}}{{a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)^{\mathrm{2}} +\left({y}−\frac{{b}^{\mathrm{2}} {c}}{{a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} \left({a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)\left(\mathrm{1}+{m}^{\mathrm{2}} \right)}{\left({a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\Rightarrow\left[\left({a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){x}+{a}^{\mathrm{2}} {mc}\right]^{\mathrm{2}} +\left[\left({a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){y}−{b}^{\mathrm{2}} {c}\right]^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} \left({a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)\left(\mathrm{1}+{m}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\left({a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\mathrm{2}\left({a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){a}^{\mathrm{2}} {mcx}+{a}^{\mathrm{4}} {m}^{\mathrm{2}} {c}^{\mathrm{2}} −\mathrm{2}\left({a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){b}^{\mathrm{2}} {cy}+{b}^{\mathrm{4}} {c}^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} \left({a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left(\mathrm{1}+{m}^{\mathrm{2}} \right)−{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} {m}^{\mathrm{2}} \\ $$$$\Rightarrow\left({a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\mathrm{2}\left({a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} {mcx}−{b}^{\mathrm{2}} {cy}\right)+\left({a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){a}^{\mathrm{2}} {c}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} \left({a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left(\mathrm{1}+{m}^{\mathrm{2}} \right)+\left({a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){b}^{\mathrm{2}} {c}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\left({a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\mathrm{2}{a}^{\mathrm{2}} {mcx}−\mathrm{2}{b}^{\mathrm{2}} {cy}+\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){c}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} \left(\mathrm{1}+{m}^{\mathrm{2}} \right)=\mathrm{0} \\ $$
Commented by ajfour last updated on 27/Jan/18
$${Thanks}\:{sir}.\:{Stay}\:{blessed}\:{Sir}. \\ $$