Question

Choose the correct options -

• A. The tangent at may point of a circle is perpendicualr to the radius through the point of contact.
• B. The lengths of two tangents drawn from an external point to a circle are equal.
• C. If all the sides of parallelogram touch a circle, the parallelogram is a rhombus.
• D. A straight line drawn at a point on the circle and perpendicular to the radius through that point is tangent to the circle.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A The lengths of two tangents drawn from an external point to a circle are equal.
B The tangent at may point of a circle is perpendicualr to the radius through the point of contact.
C If all the sides of parallelogram touch a circle, the parallelogram is a rhombus.
D A straight line drawn at a point on the circle and perpendicular to the radius through that point is tangent to the circle.

proof for A ,B,D option

At line drwn from centre to the point of intersection of tangent is always perpendicular. [shown in figure]

Here, PO and QO are radii .

So, $\angle OPT=OQT={90}^{\circ }$

Now, it is clear that both the triangles POT and QOT are right angled triangle.

and a common hypotenuse OT of these [ as shown in figure ]

Now, come to the concept ,

triangle POT and QOT

$\angle OPT=\angle OQT={90}^{\circ }$

Common hypotenuse OT

And $OP=OQ$ [ OP and OQ are radii]

So,

R - H - S rule of similarity

$\angle POT$ ~ $\angle QOT$

Hence,

$\frac{OP}{OQ}$= $\frac{PT}{QT}$ = $\frac{OT}{OT}$

$\frac{PT}{QT}=1$

$PT=QT$ [ hence proved]

proof for option C

Let ABCD be a parallelogram which circumscribes the circle.

$AP=AS$ [Tangents drawn from an external point to a circle are equal in length]

$BP=BQ$ [Tangents drawn from an external point to a circle are equal in length]

$CR=CQ$ [Tangents drawn from an external point to a circle are equal in length]

$DR=DS$ [Tangents drawn from an external point to a circle are equal in length]

Consider,

$(AP+BP)+(CR+DR)=(AS+DS)+(BQ+CQ)$

$AB+CD=AD+BC$

But $AB=CD$and $BC=AD$ [Opposite sides of parallelogram ABCD]

$AB+CD=AD+BC$

Hence$2AB=2BC$

Therefore, AB = BC

Similarly, we get

$AB=DA$ and $DA=CD$

Thus, ABCD is a rhombus.

Threfore all four are correct answer.