A circle passes through origin and meets the axes at A and B so that (2, 3) lies on AB___ then the locus of centroid of ∆OAB is

Let the circle be x2+y2+2gx+2fy+c=0 ∵ It passes through origin. ⇒c=0 ⇒x2+y2+2gx+2fy=0 When y=0, ⇒x2+2gx=0 ⇒x(x+2g)=0 ⇒x=0 or −2g ∴A=(−2g,0) When x=0, y2+2fy=0 ⇒y(y+2f)=0 ⇒y=0 or −2f ∴B=(0,−2f) Equation of AB: −x2g−y2f=1 (2,3) lies on this line. ⇒-22g−32f=1 ⇒1g+32f=−1→(1) Centroid of OAB :=(0−2g+03,0+0−2f3) ⇒(−2g3,−2f3) =(x,y) ⇒ −2g=3x ⇒g=−3x2 and −2f3=y ⇒f=−3y2 Putting in (1) ⇒−23x-33y=−1 ⇒2y+3x=3xy

A circle passes through origin and meets the axes at A and B so that (2, 3) lies on AB___ then the locus of centroid of ∆OAB is

Let the circle be x2+y2+2gx+2fy+c=0 ∵ It passes through origin. ⇒c=0 ⇒x2+y2+2gx+2fy=0 When y=0, ⇒x2+2gx=0 ⇒x(x+2g)=0 ⇒x=0 or −2g ∴A=(−2g,0) When x=0, y2+2fy=0 ⇒y(y+2f)=0 ⇒y=0 or −2f ∴B=(0,−2f) Equation of AB: −x2g−y2f=1 (2,3) lies on this line. ⇒-22g−32f=1 ⇒1g+32f=−1→(1) Centroid of OAB :=(0−2g+03,0+0−2f3) ⇒(−2g3,−2f3) =(x,y) ⇒ −2g=3x ⇒g=−3x2 and −2f3=y ⇒f=−3y2 Putting in (1) ⇒−23x-33y=−1 ⇒2y+3x=3xy

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A circle passes through origin and meets the axes at A and B so that (2, 3) lies on $\overset{___}{A B}$ then the locus of centroid of $∆$ OAB is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$3 x - 2 y = 3 x y$

$2 x - 3 y = 6 x y$

$3 x + 2 y = 3 x y$

$2 x + 3 y = 6 x y$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$L e t t h e c i r c l e b e x^{2} + y^{2} + 2 g x + 2 f y + c = 0 ∵ I t p a s s e s t h r o u g h o r i g i n . \Rightarrow c = 0 \Rightarrow x^{2} + y^{2} + 2 g x + 2 f y = 0 W h e n y = 0, \Rightarrow x^{2} + 2 g x = 0 \Rightarrow x (x + 2 g) = 0 \Rightarrow x = 0 o r - 2 g ∴ A = (- 2 g, 0) W h e n x = 0, y^{2} + 2 f y = 0 \Rightarrow y (y + 2 f) = 0 \Rightarrow y = 0 o r - 2 f ∴ B = (0, - 2 f) E q u a t i o n o f A B : \frac{- x }{2 g} - \frac{y}{2 f} = 1 (2, 3) l i e s o n t h i s l i n e . \Rightarrow \frac{- 2}{2 g} - \frac{3}{2 f} = 1 \Rightarrow \frac{1}{g} + \frac{3}{2 f} = - 1 \to (1) C e n t r o i d o f O A B : = (\frac{0 - 2 g + 0}{3} , \frac{0 + 0 - 2 f}{3} ) \Rightarrow (\frac{- 2 g}{3} , \frac{- 2 f }{3}) = (x, y) \Rightarrow - 2 g = 3 x \Rightarrow g = \frac{- 3 x }{2} a n d \frac{- 2 f }{3} = y \Rightarrow f = \frac{- 3 y }{2} P u t t i n g i n (1) \Rightarrow \frac{- 2 }{3 x} - \frac{3 }{3 y} = - 1 \Rightarrow 2 y + 3 x = 3 x y$