Find The Equation Of The Circle Concentric With The Circle x2 + y2 – 8x + 6y– 5 = 0 And Passing Through The Point (–2, –7).

  1. Home
  2. JEE Main
  3. Maths
  4. Coordinate Geometry
  5. Coordinate Systems and Coordinates

Why Kaysons ?

Video lectures

Access over 500+ hours of video lectures 24*7, covering complete syllabus for JEE preparation.

Online Support

Practice over 30000+ questions starting from basic level to JEE advance level.

Live Doubt Clearing Session

Ask your doubts live everyday Join our live doubt clearing session conducted by our experts.

National Mock Tests

Give tests to analyze your progress and evaluate where you stand in terms of your JEE preparation.

Organized Learning

Proper planning to complete syllabus is the key to get a decent rank in JEE.

Test Series/Daily assignments

Give tests to analyze your progress and evaluate where you stand in terms of your JEE preparation.

SPEAK TO COUNSELLOR ? CLICK HERE

Download JEE Question paper & Solution Download JEE Important questions Download Complete syllabus of Maths, Physics, Chemistry

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.

Download JEE Question paper & Solution Download JEE Important questions Download Complete syllabus of Maths, Physics, Chemistry

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.
medium View solution
Q2

Circum centre of the triangle PT1T2 is at
easy View solution
Q3

If P is taken to be at (h, 0) such that P’ lies on the circle, the area of the rhombus is
easy View solution
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
easy View solution
Q5

Suppose ax + bx + c = 0, where a, b, c 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
easy View solution
Q6

Find the equation of chord of x2 + y2 – 6x + 10y – 9 = 0 which is bisected at (–2, 4).
easy View solution
Q7

Find the equation of that chord of the x2 + y2 = 15 which is bisected at (3, 2).
easy View solution
Q8

 

Find the centre and radius of the circle 

             2x2 + 2y2 = 3x – 5y + 7
easy View solution
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.
easy View solution
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).
medium View solution