## 26. If the circle x² + y² + 4x + 22y + c = 0 bisects the circumference of the circle x² + y² – 2x + 8y – d = 0, then c + d is equal to

Question

26. If the circle x² + y² + 4x + 22y + c = 0 bisects the circumference of the circle
x² + y² – 2x + 8y – d = 0, then c + d is equal to :
(a) 60
(b) 50
(c) 40
(d) 30

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4 months 2022-01-06T02:15:33+00:00 2 Answers 0 views 0

=> c + d = 50

Given that, circle x² + y² + 4x + 22y + c = 0 bisects the circumference of the circle

x² + y² – 2x + 8y – d = 0

The common chord of the given circle is

S1 – S2 =0

=> x² + y² + 4x + 22y + c – x² – y² + 2x – 8y + d = 0

=> 6x + 14y + c + d = 0. ….(1)

so, Eq. (1) passes through the centre of the second circle i.e., (1, -4)

6 – 56 + c + d = 0

=> c + d = 50

Note: If S1 and S2 are the equation of the two circle, then equation of common chord is S1 – S2 = 0.

2. ## Given that, circle x² + y² + 4x + 22y + c = 0 bisects the circumference of the circle

x² + y² – 2x + 8y – d = 0

### The common chord of the given circle is

S1 – S2 =0

=> x² + y² + 4x + 22y + c – x² – y² + 2x – 8y + d = 0

=> 6x + 14y + c + d = 0. ….(1)

### so, Eq. (1) passes through the centre of the second circle i.e., (1, -4)

6 – 56 + c + d = 0

=> c + d = 50

### Note: If S1 and S2 are the equation of the two circle, then equation of common chord is S1 – S2 = 0.

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