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  • The locus of the mid-point of the chords of the circle <img alt="\mathrm{x^2+y^2-4 x-6 y-3=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5E2-4%20x-6%20y

The locus of the mid-point of the chords of the circle \mathrm{x^2+y^2-4 x-6 y-3=0} which subtend a right angle at the point \mathrm{(h, k)}, is
 

Option: 1

\mathrm{x^2+y^2+x(h-2)+y(k-3)-\frac{3}{2}+\frac{h^2+k^2}{2}=0}

 


Option: 2

\mathrm{x^2+y^2+x(h-2)-y(k-3)-\frac{3}{2}+\frac{h^2+k^2}{2}=0}
 


Option: 3

\mathrm{x^2+y^2-x(h+2)-y(k+3)-\frac{3}{2}+\frac{h^2+k^2}{2}=0}
 


Option: 4

None of these


Answers (1)

best_answer

\mathrm{ A M^2=O A^2-O M^2 }

\mathrm{ =16-\left[(p-2)^2+(q-3)^2\right] }

\mathrm{M} is midpoint of \mathrm{AB} and \mathrm{\angle A Q B=90^{\circ}}

\mathrm{ \Rightarrow AM^2=M Q^2 }

\mathrm{ 16-(p-2)^2-(q-3)^2=(h-p)^2+(k-q)^2}

Hence locus of \mathrm{(p, q)} is

\mathrm{ \mathrm{x}^2+\mathrm{y}^2-\mathrm{x}(\mathrm{h}+2)-\mathrm{y}(\mathrm{k}+3)-\frac{3}{2}+\frac{\mathrm{h}^2+\mathrm{k}^2}{2}=0 }

Hence option 3 is correct

Posted by

shivangi.shekhar

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