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  • Circle  intersect x-axis at point B and y-axis at point A then find the midpoint of AB<

Circle x^2+y^2=1 intersect x-axis at point B and y-axis at point A then find the midpoint of AB

Option: 1

(\frac{1}{2},\frac{-1}{2})


Option: 2

(\frac{1}{2},\frac{1}{2})


Option: 3

A and B both 


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=1\\ \text{so point } A(0,1) \ and B(1,0)\\ \text{mid point of AB}\\ M\equiv (\frac{0+1}{2},\frac{1+0}{2})\\ M(\frac{1}{2},\frac{1}{2})

Similiarlly it intersevt at 2nd, 3rd and 4th quardant 

Posted by

Rishi

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