Question

Prove that the line segment joining the point of contact of two parallel line to a circle is diameter of circle.

Answer 1

Answer:

Refer to the attachment is given.....

Hope it helps

Answer 2

Answer:

⋆ DIAGRAM :

$\setlength{\unitlength}{.8mm}\begin{picture}(50,55)\thicklines\qbezier(25.000,10.000)(33.284,10.000)(39.142,15.858)\qbezier(39.142,15.858)(45.000,21.716)(45.000,30.000)\qbezier(45.000,30.000)(45.000,38.284)(39.142,44.142)\qbezier(39.142,44.142)(33.284,50.000)(25.000,50.000)\qbezier(25.000,50.000)(16.716,50.000)(10.858,44.142)\qbezier(10.858,44.142)( 5.000,38.284)( 5.000,30.000)\qbezier( 5.000,30.000)( 5.000,21.716)(10.858,15.858)\qbezier(10.858,15.858)(16.716,10.000)(25.000,10.000)\put(25,10){\line(0,5){40}}\put(25,30){\circle*{1}}\put(27,29){\sf\large{O}}\put(25,10){\vector(-1,0){40}}\put(25,50){\vector(-1,0){40}}\put(25,10){\vector(1,0){40}}\put(25,50){\vector(1,0){40}}\put(23,51){\sf\large{A}}\put(49,45){\sf\large{X}}\put(50,50){\circle*{1}}\put(23,5){\sf\large{B}}\put(1,5){\sf\large{Y}}\put(2.5,10){\circle*{1}}\put(64,45){l}\put(64,12){m}\end{picture}$

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$\underline{\star\:\bold{Given :}}$

l and m are the tangents to a circle such that l || m, intersecting at point A and B respectively.

$\underline{\star\:\bold{To\:Prove :}}$

AB is a Diameter of the Circle.

$\underline{\star\:\bold{Proof :}}$

A tangent at any Point of a circle is perpendicular to the radius through the point of contact.

$\therefore\:\tt\angle\:XAO = 90^{\circ}\quad and\quad\angle\:YBO =90^{\circ}$

$\dashrightarrow\tt\:\:\angle\:XAO+\angle\:YBO= 90^{\circ}+90^{\circ}\\\\\\\dashrightarrow\tt\:\:\angle\:XAO+\angle\:YBO= 180^{\circ}$

• Angles on the same side of the transversal is 180°

∴ Line AB passes through the centre and is the diameter of the circle with radius O.