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  • The locus of the mid points of the chords of the circle <img alt="\mathrm{x^2+y^2+4 x-6 y-12=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5E2&plus;

The locus of the mid points of the chords of the circle \mathrm{x^2+y^2+4 x-6 y-12=0}  which subtend an angle of  \frac{\pi}{3} radians at its circumference is :
 

Option: 1

(\mathrm{x}-2)^2+(\mathrm{y}+3)^2=6.25


Option: 2

(\mathrm{x}+2)^2+(\mathrm{y}-3)^2=6.25


Option: 3

\mathrm{(x+2)^2+(y-3)^2=18.75}


Option: 4

\mathrm{ (x+2)^2+(y+3)^2=18.75 }


Answers (1)

best_answer

\begin{aligned} \text{radius }=5 \Rightarrow p=5 \cos 60^{\circ}=2.5\\ \Rightarrow\text{ locus is }(\mathrm{h}+2) 2+(\mathrm{k}-3) 2=6.25 \end{aligned}

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sudhir.kumar

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