Write True or False and justify your answer in each of the following :

If a number of circles touch a given line segment PQ at a point A, then their centres lie on the perpendicular bisector of PQ.

True

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False

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Ambiguous

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Data Insufficient

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Solution

The correct option is B False
$G i v e n - T w o c i r c l e s w i t h c e n t r e C & D t o u c h t h e l i n e P Q a t A . T h e r e a r e o t h e r c i r c l e s l i k e t h e s e c i r c l e s . T o f i n d o u t - T h e s t a t e m e n t t h a t c e n t r e s o f t h e c i r c l e s l i e o n t h e p e r p e n d i c u l a r b i s e c t o r o f P Q i s t r u e o r f a l s e . J u s t i f i c a t i o n - A C i s t h e l i n e j o i n i n g t h e c e n t r e t o t h e p o i n t o f c o n t a c t A o f t h e t a n g e n t P Q . ∴ A C ⊥ P Q a t A . ⟹ ∠ P A C = 90^{o} . . . . . . . . . . . (i) s i m i l a r l y, A D ⊥ P A ⟹ ∠ P A D = 90^{o} . . . . . . . . . . (i i) S o f r o m (i) & (i i) ∠ P A C + ∠ P A D = 180^{o} w h i c h i s a s t r a i g h t a n g l e . ∴ A D & A C a r e o n t h e s a m e l i n e . S o C D & P Q a r e p e r p e n d i c u l a r w i t h e a c h o t h e r a t A . N o w A i s a n y p o i n t o n P Q . ∴ P A \geq o r < \frac{1}{2} P Q . ∴ C D m a y o r m a y n o t b i s e c t P Q a t A t h o u g h P Q & C D a r e p e r p e n d i c u l a r t o e a c h o t h e r . ∴ T h e s t a t e m e n t t h a t c e n t r e s o f t h e c i r c l e s l i e o n t h e p e r p e n d i c u l a r b i s e c t o r o f P Q i s f a l s e . T h i s w i l l a p p l y f o r a n y o t h e r c i r c l e w h i c h t u o c h e s P Q a t A . A n s - F a l s e$

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