P is a variable point of the line L = 0. Tangents are drawn to the circle `x^2 + y^2 = 4` from P to touch it at Q and R. The parallelogram PQSR is completed. If `L = 2x + y - 6 = 0`, then the locus of circumcetre of `trianglePQR` is -

Options

A. `2x-y=4`
B. `2x+y=3`
C. `x-2y=4`
D. `x+2y=3`

Answer

2

Explanation

`PQ=PR`, i.e., paralleleogram PQSR is a rhombus. Therefore,
Midpoint of QR`=` Midpoint of PS and `QR _|_ PS`
Therefore , S is the mirror image of P in QR
`L -= 2x+y-6`
Let `P -= ( k,6-2k)` .
`/_ PQO = /_ PRO = (pi)/(2)`
Therefore, OP is the diameter of circumsircle PQR. Then the center is `(k//2,3-k)` . So,
`x= (k)/(2) ` or `k=2x` and `y=3-k`
Therefore, the required locus is `2x+y=3`