The point from which the tangents to the circles $x^{2} + y^{2} - 8 x + 40 = 0, 5 x^{2} + 5 y^{2} - 25 x + 80 = 0, x^{2} + y^{2} - 8 x + 16 y + 160 = 0$ are equal in length is

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If P is a point (x,y) on the line y=−3x such that P and Q(3,4) are on opposite side of the line 3x−4y=8, then:

The correct option is B x>815,y<−85y−3x....(1)3x−4y=8....(2)Put (1) in (2)3x−4(−3x)=83x+12x=815x=8x=815y=−85From the figure, it is clear thatx>815, y<−85

If $P$ is a point $(x, y)$ on the line $y = - 3 x$ such that $P$ and $Q (3, 4)$ are on opposite side of the line $3 x - 4 y = 8$ , then:

The correct option is B $x > \frac{8}{15}, y < - \frac{8}{5}$
$y - 3 x . . . . (1)$
$3 x - 4 y = 8.... (2)$
Put (1) in (2)
$3 x - 4 (- 3 x) = 8$
$3 x + 12 x = 8$
$15 x = 8$
$x = \frac{8}{15}$
$y = - \frac{8}{5}$
From the figure, it is clear that
$x > \frac{8}{15}$ , $y < - \frac{8}{5}$