Answer:
The coordinates of the centre of our circle = (0, -1)
Step-by-step explanation:
Let (x,y) represent a point on the circle.
Distance between (x,y) and B(0,3) is 2 × distance between (x,y) and O(0,0)
Distance between two points (x₁, y₁) and (x₂, y₂) in coordinates system is given as
d = √[(x₂ - x₁)² + (y₂ - y₁)²]
So, the statement from the question can be interpreted as
√[(x - 0)² + (y - 3)²] = 2 × √[(x - 0)² + (y - 0)²]
Squaring both sides
x² + (y - 3)² = 4(x² + y²)
4x² + 4y² = x² + y² - 6y + 9
3x² + 3y² + 6y - 9 = 0
Divide through by 3
x² + y² + 2y - 3 = 0
x² + y² + 2y = 3
Add 1 to both sides
x² + y² + 2y + 1 = 3 + 1
x² + y² + y + y + 1 = 4
x² + (y + 1)² = 2²
(x + 0)² + (y + 1)² = 2²
Comparing this to the equation of a circle
(x - 0)² + (y - b)² = r²
where r = radius of the circle, and the coordinates of the circle's centre is (a,b)
[x - 0]² + [y - (-1)]² = 2²
The centre of our circle = (0, -1) and radius is 2.
To check, let's see if point (√3, 0) is indeed a point on the circle. x = √3 and y = 0
x² + (y + 1)² = 2²
(√3)² + (0 + 1)² = 2²
3 + 1 = 4
4 = 4
Hence, our equation of the circle is correct and the point (√3, 0) is truly on the circle.
Hope this Helps!!!
