Question

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

Solution

Correct option is

4 (x2 – y2) = a2 – b2 

Let C (hk) be the centre of the circle passing through the end points of the rod AB and PQ of length a and b respectively, CL and CM be perpendiculars from C on AB and PQ respectively. (Fig.)

                                                                  

 

  

           PM = (1/2) PQ = b/2  

and CA = CP (radii of the same circle)

     

  

So that locus of (hk) is 4(x2 – y2) = a2 – b2

SIMILAR QUESTIONS

Q1

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

Q2

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

Q3

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  

Q4

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

Q5

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

Q6

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

Q7

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)

Q8

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

Q9

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

Q10

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