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Question Number 28512 by ajfour last updated on 26/Jan/18

Commented by ajfour last updated on 26/Jan/18

$$Ifeq.oflineABis\mathit{y}=\mathit{m}\mathit{x}+\mathit{c}$$$$andthatofellipseis\frac{{\mathit{x}}^2}{{\mathit{a}}^2}+\frac{{\mathit{y}}^2}{{\mathit{b}}^2}=1,$$$$findeq.ofcirclewithABas$$$$diameter.$$

Answered by mrW2 last updated on 26/Jan/18

$$y=mx+c$$$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$$$\frac{x^2}{a^2}+\frac{{(mx+c)}^2}{b^2}=1$$$$(a^2{m}^2+b^2)x^2+2a^{2}mcx+a^2(c^2-b^2)=0$$$$\Rightarrow x=\frac{-a^{2}mc\pm \sqrt{a^4{m}^2{c}^2-(a^2{m}^2+b^2)a^2(c^2-b^2)}}{a^2{m}^2+b^2}$$$$\Rightarrow x_{A,B}=\frac{-a^{2}mc\pm ab\sqrt{a^2{m}^2+b^2-c^2}}{a^2{m}^2+b^2}$$$$\Rightarrow x_C=\frac{x_A+x_B}{2}=-\frac{a^{2}mc}{a^2{m}^2+b^2}$$$$\Rightarrow y_C=\frac{b^{2}c}{a^2{m}^2+b^2}$$$$$$$$2R=\mid x_A-x_B\mid \sqrt{1+m^2}=\frac{2ab\sqrt{(a^2{m}^2+b^2-c^2)(1+m^2)}}{a^2{m}^2+b^2}$$$$\Rightarrow R=\frac{ab\sqrt{(a^2{m}^2+b^2-c^2)(1+m^2)}}{a^2{m}^2+b^2}$$$$$$$$Eqn.ofcircle:$$$${(x-x_C)}^2+{(y-y_C)}^2=R^2$$$$\Rightarrow {(x+\frac{a^{2}mc}{a^2{m}^2+b^2})}^2+{(y-\frac{b^{2}c}{a^2{m}^2+b^2})}^2=\frac{a^2{b}^2(a^2{m}^2+b^2-c^2)(1+m^2)}{{(a^2{m}^2+b^2)}^2}$$$$\Rightarrow {[(a^2{m}^2+b^2)x+a^{2}mc]}^2+{[(a^2{m}^2+b^2)y-b^{2}c]}^2=a^2{b}^2(a^2{m}^2+b^2-c^2)(1+m^2)$$$$\Rightarrow {(a^2{m}^2+b^2)}^2(x^2+y^2)+2(a^2{m}^2+b^2)a^{2}mcx+a^4{m}^2{c}^2-2(a^2{m}^2+b^2)b^{2}cy+b^4{c}^2=a^2{b}^2(a^2{m}^2+b^2)(1+m^2)-a^2{b}^2{c}^2-a^2{b}^2{c}^2{m}^2$$$$\Rightarrow {(a^2{m}^2+b^2)}^2(x^2+y^2)+2(a^2{m}^2+b^2)(a^{2}mcx-b^{2}cy)+(a^2{m}^2+b^2)a^2{c}^2-a^2{b}^2(a^2{m}^2+b^2)(1+m^2)+(a^2{m}^2+b^2)b^2{c}^2=0$$$$\Rightarrow (a^2{m}^2+b^2)(x^2+y^2)+2a^{2}mcx-2b^{2}cy+(a^2+b^2)c^2-a^2{b}^2(1+m^2)=0$$

Commented by ajfour last updated on 27/Jan/18

$$Thankssir.StayblessedSir.$$

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