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A variable line moves in such a way that...

A variable line moves in such a way that the product of the perpendiculars from (4, 0) and (0, 0) is equal to 9. The locus of the feet of the perpendicular from (0, 0) upon the variable line is a circle, the square of whose radius is

A

13

B

15

C

19

D

23

Text Solution

Verified by Experts

The correct Answer is:
A
Let the foot of the perpendicular from origin upon the variable line be `(x_(1),y_(1))`.
So, equation of the variable line is:
`xx_(1) +yy_(1) -(x_(1)^(2) +y_(1)^(2)) =0`

According to the equation,
`p_(1)p_(2) =|(x_(1)^(2)+y_(1)^(2))/(sqrt(x)_(1)^(2)+y_(1)^(2))||(4x_(1)-(x_(1)^(2)+y_(1)^(2)))/(sqrt(x_(1)^(2)+y_(1)^(2)))| =9`
i.e., `|x_(1)^(2) +y_(1)^(2) -4x_(1)| =9`
Locus is: `x^(2) +y^(2) - 4x -9 =0`
`:. r^(2) = 4 +9 = 13`
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