The locus of the mid-points of the chords of | vedclass

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(C) Let $(h, k)$ be the mid-point of the chord of the circle $x^2+y^2=16$. The equation of the chord with mid-point $(h, k)$ is given by $T=S_1$,which

Vedclass · Accepted Answer

$16x^2-9y^2=(x^2+y^2)^2$

Vedclass · Answer

(C) Let $(h, k)$ be the mid-point of the chord of the circle $x^2+y^2=16$. The equation of the chord with mid-point $(h, k)$ is given by $T=S_1$,which is $hx+ky=h^2+k^2$. This can be rewritten as $y = -\frac{h}{k}x + \frac{h^2+k^2}{k}$. This line is a tangent to the hyperbola $9x^2-16y^2=144$,which can be written as $\frac{x^2}{16} - \frac{y^2}{9} = 1$. 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 $c^2 = a^2m^2 - b^2$. Here $a^2=16$,$b^2=9$,$m = -\frac{h}{k}$,and $c = \frac{h^2+k^2}{k}$. Substituting these values: $\left(\frac{h^2+k^2}{k}\right)^2 = 16\left(-\frac{h}{k}\right)^2 - 9$. $\frac{(h^2+k^2)^2}{k^2} = \frac{16h^2}{k^2} - 9$. $(h^2+k^2)^2 = 16h^2 - 9k^2$. Replacing $(h, k)$ with $(x, y)$,the locus is $16x^2-9y^2=(x^2+y^2)^2$.

The locus of the mid-points of the chords of the circle $x^2+y^2=16$ which are tangents to the hyperbola $9x^2-16y^2=144$ is

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