Prove that in two concentric circle, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.

Open in App

Solution

$∠ A P O = ∠ B P O = 90^{o}$ ...Perpendicular from center to chord bisects the chord
In triangle APO,
By Pythagoras Theorem,
${O A}^{2} = {A P}^{2} + {O P}^{2}$
$∴ {A P}^{2} = {O A}^{2} -$ ${O P}^{2} (1)$
In $△ B P O$ ,
By Pythagoras Theorem,
${O B}^{2} = {B P}^{2} + {O P}^{2} ∴ {B P}^{2} = {O B}^{2} - {O P}^{2} (2)$
$O A = O B$ ...(OA and OB are radius of the same circle)
$∴$ Subtracting $(1) f r o m (2)$
${A P}^{2} = {B P}^{2}$
$∴ A P = B P$
Hence, proved.

Suggest Corrections

Similar questions

Prove that in two concentric circles, the chord of the larger circle which touches the smaller circle, is bisected at the point of contact.

Q. Prove that "In two concentric circles, a chord of the bigger circle, that touches the smaller circle is bisected at the point of contact with the smaller circle."

Q. In two concentric circles, prove that a chord of larger circle which is tangent to smaller circle is bisected at the point of contact.

In Fig.4, an isosceles triangle ABC, with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC.

OR
In Fig.5, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB.

In two concentric circles, a chord of length 8 cm of the larger circle touches the smaller circle. If the radius of the larger circle is 5 cm then find the radius of the smaller circle.

Join BYJU'S Learning Program