Question

A line is drawn through the point P(3, 11) to cut the circle x2 + y2 = 9 at A and B. Then PAPB is equal to

Solution

Correct option is

121

From geometry we know PAPB = (PT)2 where PT is the length of the tangent from P to the circle.

Hence PAPB = (3)2 + (11)2 – 9  = 112 = 121.  

SIMILAR QUESTIONS

Q1

The angle between a pair of tangents drawn from a point P to the circle x2 + y2 + 4x – 6y + 9 sin2α + 13 cos2α = 0 is 2α. The equation of the locus of the point P is

Q2

Equation of a circle through the origin and belonging to the co-axial system, of which the limiting points are (1, 2), (4, 3) is

Q3

If a line segment AM = a moves in the plane XOY remaining parallel toOX so that the left end point A slides along the circle x2 + y2 = a2, the locus of M is

Q4

If common chord of the circle C with centre at (2, 1) and radius r and the circle x2 + y2 – 2x – 6y + 6 = 0 is a diameter of the second circle, then the value of r is  

Q5

Tangents drawn from the point P(1, 8) to the circle x2 + y2 – 6x – 4y – 11 = 0 touch the circle at the points A and B. The equation of the circumcircle of the triangle in PAB is

Q6

Let ABCD be a quadrilateral with area 18, with side AB parallel to CD and AB = 2CD. Let AD be perpendicular to AB and CD. If a circle is drawn inside the quadrilateral ABCD touching all the sides, then its radius is

Q7

An equilateral triangle is inscribed in the circle x2 + y2 = a2 with the vertex at (a, 0). The equation of the side opposite to this vertex is

Q8

The lines 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154 square units. An equation of this circle is (π = 22/7)

Q9

The equation f a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is

Q10

Two rods of lengths a and b slide along the x-axid and y-axis respectively in such a manner that their ends are concyclic. The locus of the centre of the circle passing through the end points is