Question

Draw a circle with the help of a bangle. take a point outside the circle.construct the pair of tangent from this point to the circle

Answer 1

$\huge\star{\pink{\underline{\mathrm{Explanation:}}}}$

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Construction : The required tangents can be constructed on the given circle as follows.

$\sf\bold{\blue{\boxed{\underline{STEPS\:OF\:CONSTRUCTION:-}}}}$

• $\righthookarrow$Draw a circle with the help of a bangle.

• $\righthookarrow$Draw two non-parallel chords such as AB and CD

• $\righthookarrow$Draw the perpendicular bisector of AB and CD

• $\righthookarrow$Take the centre as O where the perpendicular bisector intersects.

• $\righthookarrow$To draw the tangents, take a point P outside the circle.

• $\righthookarrow$Join the points O and P.

• $\righthookarrow$Now draw the perpendicular bisector of the line PO and midpoint is taken as M

• $\righthookarrow$Take M as centre and MO as radius draw a circle.

• $\righthookarrow$Let the circle intersects intersect the circle at the points Q and R

• $\righthookarrow$Now join PQ and PR

• Therefore, PQ and PR are the required tangents.

$\huge\star{\red{\underline{\mathrm{Justification:}}}}$

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$\star$The construction can be justified by proving that PQ and PR are the tangents to the circle.

Since,O is the centre of a circle.Now, join the points OQ and OR.

(The intersection point of these perpendicular bisectors is the centre of the circle.)

Since, ∠PQO is an angle in the semi-circle.

We know that an angle in a semi-circle is a right angle.

∴ ∠PQO = 90° ⇒ OQ ⊥ PQ

Since OQ is the radius of the circle, PQ has to be a tangent of the circle.

Now, ∴ ∠PRO = 90° ⇒ or ⊥ PO Since OR is the radius of the circle.

PR has to be a tangent of the circle

Therefore, PQ and PR are the required tangents of a circle.

$\huge\star{\green{\underline{\mathrm{Hence,Proved!:}}}}$

Answer 2

Step-by-step explanation:

heey buddy

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