The equations of the tangents to the circle x | vedclass

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(A) 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,$S = x^2+y^2-4=0$ and the point is $

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$\sqrt{3} y= \pm(x-4)$

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(A) 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,$S = x^2+y^2-4=0$ and the point is $(4,0)$. $S_1 = 4^2+0^2-4 = 16-4 = 12$. $T = x(4)+y(0)-4 = 4x-4$. Substituting these into $SS_1=T^2$: $12(x^2+y^2-4) = (4x-4)^2$. $12(x^2+y^2-4) = 16(x-1)^2$. Dividing by $4$: $3(x^2+y^2-4) = 4(x^2-2x+1)$. $3x^2+3y^2-12 = 4x^2-8x+4$. $x^2-3y^2-8x+16 = 0$. $(x^2-8x+16) - 3y^2 = 0$. $(x-4)^2 - 3y^2 = 0$. $(x-4)^2 = 3y^2$. Taking the square root on both sides: $\sqrt{3}y = \pm(x-4)$.

The equations of the tangents to the circle $x^2+y^2=4$ drawn from the point $(4,0)$ are

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