Question

Passing through a point A(6, 8) a variable secant line L is drawn to the circle S: $x^2 + y^2 - 6x - 8y + 5 = 0$. From the point of intersection of L with S, a pair of tangent lines are drawn which intersect at P.

Statement-1: Locus of the point P has the equation $3x + 4y - 40 = 0$.

because

Statement-2: Point A lies outside the circle.

• A. Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
• B. Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
• C. Statement-1 is true, statement-2 is false.
• D. Statement-1 is false, statement-2 is true.

Answer 1

Answer: (D)

Statement-1 is false, statement-2 is true.

Answer 2

The equation of the circle $S$ is $x^2 + y^2 - 6x - 8y + 5 = 0$. Completing the square gives $(x-3)^2 + (y-4)^2 = 20$. The center is $C(3,4)$ and the radius is $r = \sqrt{20}$.

For Statement-2: The point $A$ is $(6, 8)$. The distance from $A$ to the center $C$ is $d = \sqrt{(6-3)^2 + (8-4)^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$. Since $d=5$ and $r=\sqrt{20}$, $d > r$, so point $A$ lies outside the circle. Statement-2 is true.

For Statement-1: The locus of point $P$ is the polar of point $A$ with respect to the circle $S$. The equation of the polar of $(x_1, y_1)$ with respect to $x^2 + y^2 + 2gx + 2fy + c = 0$ is $xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0$. Here, $(x_1, y_1) = (6, 8)$, $g=-3$, $f=-4$, $c=5$. The polar equation is $x(6) + y(8) + (-3)(x+6) + (-4)(y+8) + 5 = 0$. $6x + 8y - 3x - 18 - 4y - 32 + 5 = 0$. $3x + 4y - 45 = 0$. Statement-1 claims the locus is $3x + 4y - 40 = 0$, which is false.

Since Statement-1 is false and Statement-2 is true, the correct option is (D).