6

### Asked by gauravmore604 | 4th Nov, 2019, 03:31: PM

Expert Answer:

### (x + 2)^{2} + (y - 3)^{2} = 4
Let the coordinate be M(h, k)
B is the midpoint of A and M
B(h/2, k + 3/2)
As AB is the chord of a circle.
x^{2} + 4x + (y - 3)^{2} = 0
B must satisfy above equation
h^{2}/4 + 4h/2 + [1/2 (k + 3) - 3]^{2} = 0
h^{2} + k^{2} + 8h - 6k + 9 = 0
x^{2} + y^{2} + 8x - 6y + 9 = 0
If you simplify option B we get the same equation as above.

^{2}+ (y - 3)

^{2}= 4

^{2}+ 4x + (y - 3)

^{2}= 0

^{2}/4 + 4h/2 + [1/2 (k + 3) - 3]

^{2}= 0

^{2}+ k

^{2}+ 8h - 6k + 9 = 0

^{2}+ y

^{2}+ 8x - 6y + 9 = 0

### Answered by Sneha shidid | 5th Nov, 2019, 09:26: AM

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