In fig., circles C(O,r) and $C (O^{'}, \frac{r}{2})$ touch internally at a point A and AB is a chord of the circle C (O,r) intersecting $C (O^{'}, \frac{r}{2})$ at C. Prove that AC = CB. [3 MARKS]

Open in App

Solution

Concept: 1 Mark
Application: 2 Marks

In right triangles $Δ O C A$ and $Δ O C B$ , we have

OA = OB = r

and OC = OC

Join OA, OC and OB .

Clearly, $∠ O C A$ is the angle in a semi circle .

So, by RHS criterion of congruence, we get

$Δ O C A$ is similar to $Δ O C B$

$\Rightarrow A C = C B$

Suggest Corrections

Similar questions

Q. In the given Figure, circle

C (O, r)

and

C (O^{'}, r / 2)

touch internally at a point

A

and

A B

is a chord of the circle

C (O, r)

intersecting

C (O^{'}, r / 2)

C

. Prove that

A C = C B

Q. Two circles C (O, r) and C (O', r') intersect at two points A and B and O lies on C(O', r'). A tangent CD is drawn to the circle C(O', r') at A. Then

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.

Q. Two circles of same radii r and centres O and O' touch each other at P as shown in Fig. 10.91. If O O' is produced to meet the cirele C (O', r) at A and AT is a tangent to the circle C (O,r) such that O'Q

⊥

AT. Then AO : AO' =

(a) 3/2

(b) 2

(c) 3

(d) 1/4

Two circles touch each other internally at a point P. A chord AB of the bigger circle intersects the other circle in C and D. Prove that : $∠ C P A = ∠ D P B$ .

Join BYJU'S Learning Program

In fig., circles C(O,r) and C(O′,r2) touch internally at a point A and AB is a chord of the circle C (O,r) intersecting C(O′,r2) at C. Prove that AC = CB. [3 MARKS]

Concept: 1 Mark Application: 2 Marks In right triangles ΔOCA and ΔOCB, we have OA = OB = r and OC = OC Join OA, OC and OB . Clearly, ∠OCA is the angle in a semi circle . So, by RHS criterion of congruence, we get ΔOCA is similar to ΔOCB ⇒AC=CB

In fig., circles C(O,r) and $C (O^{'}, \frac{r}{2})$ touch internally at a point A and AB is a chord of the circle C (O,r) intersecting $C (O^{'}, \frac{r}{2})$ at C. Prove that AC = CB. [3 MARKS]