Question

- Prove that the perpendicular at a point of contact to the tangent to a circle passes through the centre

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Answer 1

Step-by-step explanation:

Given:

• A circle with tangent and centre O.

To Prove:

• Perpendicular passes through the centre.

Proof: Let PT be a tangent to the circle and P is the point of contact.

Let PQ ⟂ PT , where Q lies on the circle.

Therefore,

• ∠QPT = 90°......(1)

Let again PQ not pass through the centre O. Join PO as the radius of the circle through the point of contact.

Now we have

• PO ⟂ PT {tangent is perpendicular to the radius through the point of contact}

Therefore, ∠OPT = 90° or ∠RPT = 90°.....(2)

From equation (1) & (2) we got

∠QPT = ∠RPT = 90°

But this is possible only if P , Q and R are collinear. But a straight line cuts a circle in at the most two points.

∴ Points Q & R must coincide.

Hence, PQ passes through the centre of circle.

$\large\bold{\texttt {Proved }}$

Answer 2

✼Given:-

➣ A circle with tangent and centre O.

✼To Prove:-

➣ perpendicular at a point of contact to the tangent to a circle passes through the centre.

✼SOLUTION:-

➨ Let AB be the tangent drawn at the point P on the circle with centre O.

➨ If possible, let PQ be perpendicular to AB, not passing through O.

➨ Join OP.

➨ Since tangent at a point to a circle is perpendicular to the radius through the point, therefore

$\sf \longrightarrow AB \perp OP, i.e., \angle OPB = 90\degree$

$\sf \longrightarrow Also, \angle QPB = 90\degree (construction)$

$\sf \therefore \angle QPB = \angle OPB, which \: is \: not \: possible \\ \sf because \: as \: a \: part \: cannot \: be \: equal \: to \\ \sf the \: whole.$

$\sf ➨ Thus, it \: contradicts \: our \: supposition.$

$\sf➲ Hence,the \: perpendicular \: at \: a \: point \: of \\ \sf contact \: to \: the \: tangent \: to \: a \: circle \: passes \\ \sf through \: the \: centre.</strong><strong>$

$\huge{ \bold{ Hence, Proved }}$