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Question Number 36880 by Tinkutara last updated on 06/Jun/18

Commented by ajfour last updated on 07/Jun/18

Commented by ajfour last updated on 07/Jun/18

$$2h=a\cos \theta +b\sin \theta$$$$2k=b\cos \theta +a\sin \theta$$$$\Rightarrow \cos \theta =\frac{2(ah-bk)}{a^2-b^2}$$$$\sin \theta =\frac{2(bh-ak)}{a^2-b^2}$$$$solocuseq.is$$$${(ax-by)}^2+{(bx-ay)}^2=\frac{{(a^2-b^2)}^2}{4}.$$$${doesn}^{\prime }tmatchwithoptions..?!$$

Commented by Tinkutara last updated on 08/Jun/18

Sir please explain first 2 lines.

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jun/18

$$letrodlength\mathit{a}slideonaxes.itslideat(x_1,0)$$$$and(0,y_1)atanyinstant$$$$so{(x_1-0)}^2+{(0-y_1)}^2=a^2$$$$similarly{\left(x_2\right)}^2+{\left(y_2\right)}^2=b^2$$$$letthecentreofcircleis(h,k)$$$$since{(x-h)}^2+{(y-k)}^2=r^{2}passesthrough.{.}^{}$$$$(x_1,0),(0,y_1),(x_2,0),(0,y_2)$$$${(x_1-h)}^2+{(0-k)}^2=r^2$$$${(0-h)}^2+{(y_1-k)}^2=r^2$$$${(x_2-h)}^2+{(0-k)}^2=r^2$$$${(0-h)}^2+{(y_2-k)}^2=r^2$$$${(x_1-h)}^2={(x_2-h)}^2$$$$x_1^2-2{hx}_1=x_2^2-2{hx}_2$$$$\text{Missing left or extra right}$$$$(x_1-x_2)(x_1+x_2-2h)=0$$$$h=\frac{x_1+x_2}{2}$$$$h^2+{(y_1-k)}^2=h^2+{(y_2-k)}^2$$$${(y_1-k)}^2-{(y_2-k)}^2=0$$$$y_1^2-2y_{1}k-y_2^2+2y_{2}k=0$$$$(y_1+y_2)(y_1-y_2)-2k(y_1-y_2)=0$$$$k=\frac{y_1+y_2}{2}$$$$x_1^2+y_1^2=a^2$$$$x_2^2+y_2^2=b^2$$$${(x_1+x_2)}^2=4h^2$$$$x_1^2+x_2^2+2x_1{x}_2=4h^2$$$$y_1^2+y_2^2+2y_1{y}_2=4k^2$$$$a^2+b^2+2(x_1{x}_2+y_1{y}_2)=4(h^2+k^2)$$$$contd$$$$$$$$$$$$$$

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