To the circle `x^2 + y^2 = 4` two tangents are drawn from `P (-4, 0)`, which touch the circle at `T_1`, and `T_2` and a rhombus `PT_1P'T_2` is completed. Circumcentre of the triangle `PT_1T_2` is at

Options

A. `(-2,0)`
B. `(2,0)`
C. `(sqrt(3)//2,0)`
D. none of these

Answer

1

Explanation

`PT _(2)=PT_(1)=sqrt((-4)^(2)+o^(2)-4)=2sqrt(3)`
The circumcenter of triagnel `PT_(1)T_(2)`is the mdipoint of PO as
`/_PT_(1)O =/_PT_(2)O=90^(@)`
so, `((-4+0)/(2),(0+0)/(2))-=(-2,0)`