The centre of the circle passing through (0,0) and (1,0) and touching the circle x^2+y^2=9 is

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asked Sep 22, 2017 in Mathematics by ankitdixit (1,339 points) 13 60

The centre of the circle passing through (0,0) and (1,0) and touching the circle x2+y2=9 is

1 Answer

+2 votes
answered Sep 22, 2017 by anukriti (13,536 points) 5 10 42
selected Sep 25, 2017 by sarthaks
 
Best answer

We have General Equation of Circle is given as 

x2+y2+2gx+2fy+c=0 ---(1)

Since, Circle passes through, (0,0) and (1,0)

Therefore Equation (1) reduces to,

c=0 ---(2)

1+2g+c=0 ---(3),

From (2) and (3), g= -1/2

Also, From As per Question, 

Cicle in equation, (1) touches, another circle having equation, 

x2+y2=(3)2 ---(4)

SInce the center (0,0) of circle (4) lies on the loci of the circle (1), therefore, The radius of the circle (4) must be equal to diameter of circle(1)

hence, 2√g2+f2-c= 3

=> √1/4+f2=3/2

=> f2= 9/4-1/4

=> f2= 2

=> f= ±√2

Hence Centres of Circle (1) is (-g,-f) or (1/2,±√2)

Option (B) is correct

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