The locus of the midpoints of the chord of th | vedclass

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(D) Let the midpoint of the chord be $(h, k)$.The equation of the chord of the circle $x^{2}+y^{2}=25$ with midpoint $(h, k)$ is given by $T=S_1$,whic

Vedclass · Accepted Answer

$\left(x^{2}+y^{2}\right)^{2}-9x^{2}+16y^{2}=0$

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(D) Let the midpoint of the chord be $(h, k)$.The equation of the chord of the circle $x^{2}+y^{2}=25$ with midpoint $(h, k)$ is given by $T=S_1$,which is $xh+yk=h^{2}+k^{2}$.This can be rewritten as $y = -\frac{h}{k}x + \frac{h^{2}+k^{2}}{k}$.This line is tangent to the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$.The condition for the line $y=mx+c$ to be a tangent to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is $c^{2}=a^{2}m^{2}-b^{2}$.Here,$m=-\frac{h}{k}$ and $c=\frac{h^{2}+k^{2}}{k}$,$a^{2}=9$,$b^{2}=16$.Substituting these values into the condition: $\left(\frac{h^{2}+k^{2}}{k}\right)^{2} = 9\left(-\frac{h}{k}\right)^{2} - 16$.$\frac{(h^{2}+k^{2})^{2}}{k^{2}} = \frac{9h^{2}}{k^{2}} - 16$.$(h^{2}+k^{2})^{2} = 9h^{2} - 16k^{2}$.Replacing $(h, k)$ with $(x, y)$,the locus is $(x^{2}+y^{2})^{2} = 9x^{2}-16y^{2}$,or $(x^{2}+y^{2})^{2}-9x^{2}+16y^{2}=0$.

The locus of the midpoints of the chord of the circle $x^{2}+y^{2}=25$ which is tangent to the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$ is

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