The locus of the middle points of chords of t | vedclass

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(D) Let the midpoint of the chord be $(h, k)$.The equation of the chord with midpoint $(h, k)$ is given by $T = S_1$,where $T$ is the tangent-like exp

Vedclass · Accepted Answer

$x^2 + y^2 - x - 3y = 0$

Vedclass · Answer

(D) Let the midpoint of the chord be $(h, k)$.The equation of the chord with midpoint $(h, k)$ is given by $T = S_1$,where $T$ is the tangent-like expression and $S_1$ is the value of the circle equation at $(h, k)$.The equation of the circle is $S: x^2 + y^2 - 2x - 6y - 10 = 0$.Thus,$hx + ky - (x + h) - 3(y + k) - 10 = h^2 + k^2 - 2h - 6k - 10$.Since the chord passes through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$ into the equation:$h(0) + k(0) - (0 + h) - 3(0 + k) - 10 = h^2 + k^2 - 2h - 6k - 10$.$-h - 3k - 10 = h^2 + k^2 - 2h - 6k - 10$.Simplifying the equation:$h^2 + k^2 - h - 3k = 0$.Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 - x - 3y = 0$.

The locus of the middle points of chords of the circle $x^2 + y^2 - 2x - 6y - 10 = 0$ which pass through the origin is:

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