You visited us 1 times! Enjoying our articles? Unlock Full Access!

Standard Equation of Ellipse

In figure O...

Question

In figure $O$ is the centre of the circle. $P Q$ is tangent to the circle and secant $P A B$ passes through the centre $O$ . If $P Q = 5 c m$ and $P A = 1 c m$ , then the radius of the circle is

24 c m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10 c m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

12 c m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

5 c m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $12 c m$
$G i v e n - T h e s e c a n t ¯ P O B & t h e t a n g e n t P Q t o a c i r c l e w i t h c e n t r e O h a v e b e e n d r a w n f r o m a p o i n t P o u t s i d e t h e c i r c l e . P Q t o u c h e s t h e c i r c l e a t Q a n d ¯ P O B i n t e r s e c t s t h e t h e c i r c l e a t A & B . P Q = 5 c m & P A = 1 c m . T o f i n d o u t - t h e r a d i u s o f t h e c i r c l e = x = ? S o l u t i o n - ¯ A O B i s t h e d i a m e t e r o f t h e c i r c l e s i n c e ¯ P O B p a s s e s t h r o u g h O . N o w P A i s t h e s e g m e n t o f t h e s e c a n t ¯ P O B o u t s i d e t h e c i r c l e . S o, b y t a n g e n t - s e c a n t r u l e, t h e s q u a r e o f t h e t a n g e n t = s e c a n t s e g m e n t \times s e g m e n t o f t h e s e c a n t o u t s i d e t h e c i r c l e w h e n a s e c a n t a n d t a n g e n t t o a c i r c l e a r e d r a w n f r o m a p o i n t o u t s i d e t h e c i r c l e . i . e P B \times P A = {P Q}^{2} ⟹ P B = \frac{{P Q}^{2}}{P A} = \frac{25^{2}}{1} c m = 25 c m . ∴ T h e d i a m e t e r ¯ A O B = P B - P A = (25 - 1) c m = 24 c m . ∴ T h e r a d i u s = x = \frac{¯ A O B}{2} = \frac{24}{2} c m = 12 c m . A n s - O p t i o n C .$

Suggest Corrections

Similar questions

Q. A tangent

P Q

at a point

P

of a circle of radius

5 c m

meets a line through the centre

O

at a point

Q

so that

O Q = 12 c m

. Length of

P Q

O

is the centre of circle.

R Q

is tangent to circle at

Q

and

R P

is tangent to circle at

P

O P = 5 c m

P Q = 8 c m

O M ⊥ P Q

Find

M Q

Q. In the figure O is the centre of the circle of radius

10 c m

O P ⊥ A B, O Q ⊥ C D, ¯ ¯¯¯¯¯¯¯ ¯ A D ¯ ¯¯¯¯¯¯¯ ¯ C D, A B = 12 c m, A B = 12 c m a n d C D = 16 c m

then PQ=____

Q. In the figure, the radii of two concentric circles with centre

O

are

5 c m

and

3 c m

P A

and

P B

are tangents to these circles. If

P A = 12 c m

. Find

B P

In the given figure, O is the centre of two concentric circles fo radii 5 cm and 3 cm. From an external point P tangents PA and PB are drawn to these circles. If PA = 12 cm then PB is equal to

(a) $5 \sqrt{2}$ cm (b) $3 \sqrt{5}$ cm (c) $4 \sqrt{10}$ cm (d) $5 \sqrt{10}$ cm

Join BYJU'S Learning Program

In figure O is the centre of the circle. PQ is tangent to the circle and secant PAB passes through the centre O. If PQ=5cm and PA=1cm, then the radius of the circle is

In figure $O$ is the centre of the circle. $P Q$ is tangent to the circle and secant $P A B$ passes through the centre $O$ . If $P Q = 5 c m$ and $P A = 1 c m$ , then the radius of the circle is