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A circle passes through (0,0) and (1, 0)...

A circle passes through `(0,0)` and `(1, 0)` and touches the circle `x^2 + y^2 = 9` then the centre of circle is -

A

(3/2, 1/2)

B

(1/2, 3/2)

C

(1/2, 1/2)

D

`(1/2, -2^(uu2))`

Text Solution

Verified by Experts

The correct Answer is:
D
Let `C_(1)(h,k)` be the centre of the required circle. Then,
`sqrt((h-0)^(2)+(k-0)^(2))=sqrt((h-1)^(2)+(k-0)^(2))`
`rArr h^(2)+k^(2)=h^(2)-2h+1+k^(2)`
`rArr -2h+1=0 rArr h=1//2`
Since, (0, 0) and(1, 0) lie inside the circle `x^(2)+y^(2)=9` Therefore, the required circle can touch the given circle internally.
i.e. `C_(1)*C_(2)=r_(1)~r_(2)`
`rArr sqrt(h^(2)+k^(2))=3-sqrt(h^(2)+k^(2))`
`rArr 2sqrt(h^(2)+k^(2))=(3)/(2) rArr2sqrt((1)/(4)+k^(2))=3`
`rArr sqrt((1)/(4)+k^(2))=(3)/(2)rArr (1)/(4)+k^(2)=(9)/(4)`
`rArr k^(2)=2 rArr k=pmsqrt2`
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