The locus of the centre of a circle touching the circle `x^2 + y^2 - 4y -2x = 2sqrt3 - 1` internally and tangents on which from (1,2) is making a `60^@` angle with each other is a circle. then integral part of its radius is

Options

A. `(x-1)^(2)+(y-2)^(2)=3`
B. `(x-2)^(2)+(y-1)^(2)=1+2sqrt(3)`
C. `x^(2)+y^(2)=1`
D. none of these

Answer

D

Explanation

Let `C_(2) (h, k) `be the centre of the circle touching the given circle internally and r be its radius.

We have,
`C_(1)P=sqrt(1+4+2sqrt(3)-1)=sqrt(4+2sqrt(3))`
Since the circles touch each other internally.
`:. C_(1)C_(2)=C_(1)P-C_(2)P`
`rArr sqrt((h-1)^(2)+(k-2)^(2))=sqrt(4+2sqrt(3))-r`
`rArr r=sqrt(4+2sqrt(3))-sqrt((h-1)^(2)+(k-2)^(2))`
In `Delta C_(1) MC_(2)`, we have
`sin 30^(@)=(C_(2)M)/(C_(1)C_(2))`
`rArr (1)/(2)=(sqrt(4+2sqrt(3))-sqrt((h-1)^(2)+(k-2)^(2)))/(sqrt((h-1)^(2)+(k-2)^(2)))`
`rArr 3sqrt((h-1)^(2)+(k-2)^(2))=2sqrt(4+2sqrt(3))`
`rArr 9[(h-1)^(2)+(k-2)^(2)]=4(4+2sqrt(3))`
Hence, the locus of (h, k) is
`9[(x-1)^(2)+(y-2)^(2)]=4(4+2sqrt(3))`