The equations of the tangents drawn from the | vedclass

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Question

(B) The equation of the pair of tangents from a point $(x_1, y_1)$ to a circle $S = 0$ is given by $SS_1 = T^2$.Here,the point is $(0, 0)$,so $S_1 = 0

Vedclass · Accepted Answer

$({h^2} - {r^2})x - 2rhy = 0, x = 0$

Vedclass · Answer

(B) The equation of the pair of tangents from a point $(x_1, y_1)$ to a circle $S = 0$ is given by $SS_1 = T^2$.Here,the point is $(0, 0)$,so $S_1 = 0^2 + 0^2 - 2r(0) - 2h(0) + h^2 = h^2$.The equation of the tangent $T$ at $(0, 0)$ is $x(0) + y(0) - r(x + 0) - h(y + 0) + h^2 = 0$,which simplifies to $T = -rx - hy + h^2$.Substituting these into $SS_1 = T^2$:$h^2(x^2 + y^2 - 2rx - 2hy + h^2) = (-rx - hy + h^2)^2$.Expanding both sides:$h^2x^2 + h^2y^2 - 2rh^2x - 2h^3y + h^4 = r^2x^2 + h^2y^2 + h^4 + 2rhxy - 2rh^2x - 2h^3y$.Canceling common terms $h^2y^2, h^4, -2rh^2x, -2h^3y$:$h^2x^2 = r^2x^2 + 2rhxy$.$(h^2 - r^2)x^2 - 2rhxy = 0$.$x((h^2 - r^2)x - 2rhy) = 0$.Thus,the equations are $x = 0$ and $(h^2 - r^2)x - 2rhy = 0$.

The equations of the tangents drawn from the origin to the circle ${x^2} + {y^2} - 2rx - 2hy + {h^2} = 0$ are

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