Question

Suppose ax + bx + c = 0, where abc are in A.P. be normal to a family or circles. The equation of the circle of the family which intersects the circle x2 + y2 – 4x – 4y – 1 = 0 orthogonally is

Solution

Correct option is

x2 + y2 – 2x + 4– 3 = 0

 

abc are in A.P., so ax + by + c = 0 represents a family of lines passing through the point (1, –2). So, the family of circles (concentric) will be given by x2 + y2 – 2x + 4y + c = 0.

And 

SIMILAR QUESTIONS

Q1

Find the point of intersection of the line 2x + 3y = 18 and the circle x2 +y2 = 25.

Q2

Find the equation of the normal to the circle x2 + y2 – 5x + 2y – 48 = 0 at point (5, 6).

Q3

Find the equation of the tangent to the circle x2 + y2 = 16 drawn from the point (1, 4).

Q4

Find the equation of the image of the circle x2 + y2 + 16x – 24y + 183 = 0 by the line mirror 4x + 7y + 13 = 0.

Q5

Find the equation of the normal to the circle x2 + y2 = 2x, which is parallel to the line x + 2y = 3.

Q6

 

Find the equation of the circle which cuts orthogonally each of the three circles given below: 

x2 + y2 – 2x + 3y – 7 = 0, x2 + y2 + 5x – 5y + 9 = 0 and x2 + y2 + 7x – 9x + 29 = 0.

Q7

Circum centre of the triangle PT1T2 is at

Q8

If P is taken to be at (h, 0) such that P’ lies on the circle, the area of the rhombus is

Q9

Locus of mid-point of the chords of contact of x2 + y2 = 2 from the points on the line 3x + 4y = 10 is a circle with centre P. If O be the origin then OP is equal to

Q10

Find the equation of chord of x2 + y2 – 6x + 10y – 9 = 0 which is bisected at (–2, 4).