Question

show that the points (11,2) is the centre of the circle passing through the point (1,2) and (3,-4)​

Answer 1

Answer:

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Step-by-step explanation:

I think so there must be a diagram....

Answer 2

Answer:

Let equation of general equation of circle is x² + y² + 2gx + 2fy + c = 0 then, centre of circle will be (-g , - f)

Circle passing through three points (1,2) , (3,-4) and (5,-6) .

Put (1,2) in equation of circle.

1² + 2² + 2g(1) + 2f(2) + c = 0

2g + 4f + c + 5 = 0 -----(1)

similarly, put (3,-4)

3² + (-4)² + 2g(3) + 2f(-4) + c = 0

9 + 16 + 6g - 8f + c = 0

6g - 8f + c + 25 = 0 ------(2)

put (5, -6),

5² + (-6)² + 2g(5) + 2f(-6) + c = 0

25 + 36 + 10g - 12f + c = 0

10g - 12f + c + 51 = 0 ------(3)

subtracting equation (1) from (2),

6g - 8f + c + 25 - 2g - 4f - c - 5 = 0

4g - 12f + 20 = 0

g - 3f + 5 = 0 -------(4)

subtracting equation (2) from (3),

10g - 12f + c + 51 - 6g + 8f - c - 25 = 0

4g - 4f + 36 = 0

g - f + 9 = 0 -------(5)

solve equations (4) and (5),

f - 9 -3f + 5 = 0

-2f - 4 = 0 ⇒ f = -2

And g = f - 9 = -11

Put g and f in equation (1),

-22 - 8 + c + 5 = 0

⇒ c = 25

Hence, equation of circle is x² + y² -22x -4x + 25 = 0

Centre of circle , C ≡ (- g, - f) ≡ (11, 2)

Hence, it is proved that (11,2) is centre of circle passing through the points (1,2), (3,-4) and (5,-6)

Step-by-step explanation: