Find The Radical Centre Of The Circles, x2 + y2 + 3x + 2y + 1 = 0,  x2 + y2 – x + 6y + 5 = 0, x2 + y2 + 5x – 8y + 15 = 0

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Question

Find the radical centre of the circles, x2 + y2 + 3x + 2y + 1 = 0,  x2 + y2 – x + 6y + 5 = 0, x2 + y2 + 5x – 8y + 15 = 0

Solution

Correct option is

(3, 2)

Given circle are 

                   x2 + y2 + 3x + 2y + 1 = 0                    …(1)  

                   x2 + y2 – x + 6y + 5 = 0                      …(2)  

                   x‑2 + y2 + 5x – 8y + 15 = 0                 …(3)  Equation (1) – (2)  

                   4x – 4y – 4 = 0                                   …(4)

Equation (2) – (3)  

                     

                     

Equation (1) – (3) 

                     

                     

Equation (4) + (6)  

                   4y = 8  

                     y = 2  

From (4)

                   x – 2 = 1  

                         x = 3  

So the radical centre of the circle is (3, 2).

SIMILAR QUESTIONS

Q1

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Q2

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Q3

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Q4

Let a circle be given by 2x(x – a) + y(2y – b) = 0, (a  0, b ≠ 0). Find the condition on a and b, if two chords each bisected by the x – axis can be drawn to the circle from 

Q5

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Q6

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Q7

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Q8

The normal 3x – 4y = 4 and 6x – 8y – 7 = 0 are tangents to the circle. Then its radius is:

Q9

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Q10

The tangents drawn from the origin to the circle x2 + y2 – 2kx – 2ry + r2 = 0 are perpendicular, if: