LOCuuuus

find locus of center of circle which touches (y-1)2+x2=1 externally and also touches X axis.

13 Answers

33
Abhishek Priyam ·

(k-1)2+h2=(k+1)2
giving

x2=4y
also x=0

so 2 curves... are there....

11
Mani Pal Singh ·

abhishek
yeh h , k kya liye hain???

24
eureka123 ·

samajh nahijn aaya........[2][2]

33
Abhishek Priyam ·

(h,k) center of circle..

since it touches x axis so radius is k

distance from 0,1 =1+k (sum of radii)

1
vector ·

priyam is right

11
Mani Pal Singh ·

abhishek i diasgree


(k-1)2+h2=(k+1)2
giving

x2=4y
also x=0

so 2 curves... are there....

IF (h,k) IS THE CENTER OF THE NEW CIRCLE THEN HOW COME THIS EQUATION FORMED????

1
vector ·

diagram banao samaj me aa jayega
lhs is distance formula bw new circle centre n old one centre
rhs is k+1
k=radius of new n 1 radius of old

11
Mani Pal Singh ·

if (h,k )is the center then how would it satisfy the equation (y-1)2+x2=1
i show u the diagram

1
vector ·

h,k is nt satisfying the eqn its the centre of reqd circle

33
Abhishek Priyam ·

(y-1)2+x2=(y+1)2

11
Mani Pal Singh ·

abhishek in #11u r saying that the center is fixed at (0,1) but the radius varies with y

please make me understand this thing

1
vector ·

ya even me got the same ans

1
vector ·

we r asked the centre of circle rite... so priyam assumed it to be (h,k).as the circle is touching x axis so its centre y coordinate tht s k =radius .now after that he used the 2nd cndition that s tangent to given circle having centre at (0,1) n radius 1 n thus distance bw centres is equal to 1+k(r1+r2)
r u cnvinced now/

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