Question

Two points P and Q are taken on the line joining the points A (0, 0) and B (3a, 0) such that AP = PQ = QB. Circles are drawn onAPPQ, and QB as diameters. The locus of the point S, the sum of the squares of the lengths of the tangents from which to the three circles is equal to b2, is

Solution

Correct option is

3(x2 + y2) – 9ax + 8a2 – b2 = 0 

Since AP = PQ = QB. The coordinates of P are (a, 0) and of Q are (2a, 0). Equations of the circles on APPQ and QB as diameters are respectively.

(x – 0) (x – a) + y2 = 0, (x – a) (x – 2a) + y2 = 0 and  

(x – 2a) (x – 3a) + y2 = 0

So, if (hk) be any point on the locus, then  

        3(h2k2) – 9ah + 8a2 = b2    

So the required locus of (hk) is

        3(x2 + y2) – 9ax + 8a2 – b2 = 0.

                                                                              

SIMILAR QUESTIONS

Q1

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

Q2

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

Q3

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

Q4

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)

Q5

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

Q6

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

Q7

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

Q8

If the point (1, 4) lies inside the circle x2 + y2 – 6x – 10y + p = 0 and the circle does not touch or interest the coordinates axes, then

Q9

If the line x cos α + y sin α = p represents the common chord APQB of the circle x2 + y2 = a2 and x2 + y2 = b2 (a > b) as shown in the Fig, then AP is equal to

Q10

If OA and OB are two equal chords of the circle x2 + y2 – 2x + 4y = 0 perpendicular to each other and passing through the origin O, the slopes of OA and OB are the roots of the equation