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  • AB is a diameter of a circle; CD is a chord parallel to AB and

AB is a diameter of a circle; CD is a chord parallel to AB and \mathrm{2 C D=A B}. The tangent at B meets the line AC produced at E. Then \mathrm{A E=\lambda A B}. The value of \lambda is ___________.

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

Let the circle be x^2+y^2=a^2 and the diameter AB be chosen along x-axis so that A is (a, 0) and B is (-a, 0). Since CD is parallel to AB and half of its length therefore x-co-ordinate of C will be a / 2 and that of D will be -a / 2.

\therefore \quad C \text { is }\left(\frac{a}{2},-\frac{\sqrt{3}}{2} a\right) .
Equation of AC is y-0=\sqrt{3}(x-a).             ....(i)

Tangent at B is x=-a                                          ....(ii)

(i) and (ii) meet at E whose co-ordinates are (-a,-2 \sqrt{3} a)

\begin{array}{ll} \therefore & A E^2=(a+a)^2+(2 \sqrt{3} a)^2=4 a^2+12 a^2=16 a^2 \\ \therefore & A E=4 a=2 A B . \end{array}

Posted by

Sumit Saini

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