Question

Find the equation of the circle concentric with the circle x2 + y2 – 8x + 6y– 5 = 0 and passing through the point (–2, –7).

Solution

Correct option is

x2 + y2 – 8x + 6y – 27 = 0

 

The given equation of circle is

              x2 + y2 – 8x + 6y – 5 = 0 

Therefore, the centre of the circle is at (4, –3). Since the required circle is concentric with this circle, therefore, the centre of the required circle is also at (4, –3). Since the point (–2, –7) lies on the circle, the distance of the centre from this point is the radius of the circle. Therefore, we get

                

Hence, the equation of the circle becomes

             (x – 4)2 + (y + 3)2 = 52

or          x2 + y2 – 8x + 6y – 27 = 0.

SIMILAR QUESTIONS

Q1

 

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.

Q2

Circum centre of the triangle PT1T2 is at

Q3

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

Q4

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

Q5

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

Q6

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

Q7

Find the equation of that chord of the x2 + y2 = 15 which is bisected at (3, 2).

Q8

 

Find the centre and radius of the circle 

             2x2 + 2y2 = 3x – 5y + 7

Q9

Find the equation of the circle whose centre is the point of intersection of the lines 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 and passes through the origin.

Q10

A circle has radius 3 units and its centre lies on the line y = x – 1. Find the equation of the circle if it passes through (7, 3).