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Circle x^2+y^2-2x-2y+1=0 has a  chord AB=\sqrt{2} then find the Locus of the midpoint of chord AB

Option: 1

(x-1)^2+(y-1)^2=\frac{1}{2}


Option: 2

(x-1)^2+(y+1)^2=\frac{1}{2}


Option: 3

(x+1)^2+(y+1)^2=\frac{1}{2}


Option: 4

None of these


Answers (1)

best_answer

 

 

Locus of Mid Point of the Chord of the Circle -

Locus of Mid Point of the Chord of the Circle

A circle with radius r,  centered at the point (h, k) and AB is its chord. Let M (x1 , y1) be the midpoint of the chord AB. 

From the figure,
\\\mathrm{\cos\theta=\frac{CM}{\mathit{r}}=\frac{\sqrt{(x_1-h)^2+(y_1-k)^2}}{\mathit{r}}}\\\\\mathrm{\Rightarrow 1-\sin^2\theta=\frac{{(x_1-h)^2+(y_1-k)^2}}{\mathit{r^2}}}\\\\\mathrm{\Rightarrow \;\;\;-\sin^2\theta=\frac{{(x_1-h)^2+(y_1-k)^2}-r^2}{\mathit{r^2}}}\\\\\therefore \mathrm{Required\;equation\;of\;locus\;is}\\\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\frac{{(x-h)^2+(y-k)^2}-r^2}{\mathit{r^2}}=-\sin^2\theta}

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x^2+y^2-2x-2y+1=0\\ (x-1)^2+(y-1)^2=1\\ \text{circle with centre (1,1) and radius r=1}\\

\text{From image you can see that mid point of AB also make a circle with centre (1,1)}\\ \text{length of AB }=\sqrt{2}\\ \text{take point }A(1,0),B(0,1)\\ M \equiv (\frac{1}{2}.\frac{1}{2})\\ \text{Radius of small circle }=\sqrt{(\frac{1}{2})^2+(\frac{1}{2})^2}=\frac{1}{\sqrt{2}}\\ \text{Locus of midpoint} \Rightarrow (x-1)^2+(y-1)^2=(\frac{1}{\sqrt{2}})^2=\frac{1}{2}

Posted by

Pankaj Sanodiya

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