If a circle passing trough the point (-1, 0) touches y - axis (0,2) , then the length of the chord of the circle along the x - axis is:

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Solution

$\begin{matrix} C i r c l e t o u c h i n g y - a x i s a n d h a v i n g r a d i u s r x^{2} + y^{2} + 2 g x \pm 2 r y + g^{2} = 0 N o w, e q u a t i o n o f a c i r c l e {(x - h)}^{2} + {(y - k)}^{2} = r^{2} w h e r e, c i r c l e w i t h c e n t r e (h, k) a n d r a d i u s r . I f t h e c e n t r e i s (h, 2) t h e n r a d i u s = | h | e q u a t i o n o f c i r c l e i s {(x - h)}^{2} + {(y - 2)}^{2} = h^{2} a n d i t p a s s e s t h r o u g h p o i n t (- 1, 0) p u t t i n g v a l u e, w e g e t h = \frac{- 5}{2} S o, c e n t r e i s (\frac{- 5}{2}, 2) H e n c e, e q u a t i o n {(x + \frac{5}{2})}^{2} + {(y - 2)}^{2} = {(\frac{5}{2})}^{2} (- 4, 0) s a t i s f i e s t h i s e q u a t i o n . \end{matrix}$

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