The equation of a circle is given by + y² = a where a is radius. If the equation is modified to change the origin other then (0, 0). Then find out the correct dimensions of A and B in a new equation (x-4)²-(y-2) = 0 The dimension of t is given as [7¹] 1) A = [LT], B = [LT] 3) A = [2¹T¹,8-[LT] 2) A = [LT], B =[['T] 4) A=[LT].B=[LT]
Answers & Comments
Answer:
(x - At)2 + (y - t/B)2 = a2. The dimensions of t is given as [T–1 ]. (1) A = [L–1 T], B = [LT–1 ] (2) A = (LT], B = [L–1T–1 ] (3) A = [L–1T–1 ], B = [LT–1 ] (4) A = [L–1T–1 ], B = [LT]
Answer:
The equation of a circle in standard form is given by:
(x - h)² + (y - k)² = r²
where (h, k) are the coordinates of the center of the circle, and r is the radius.
In the modified equation, (x-4)² + (y-2) = 0, we can compare it with the standard form to find the new coordinates of the center of the circle.
Comparing the equations:
(x - h)² + (y - k)² = r²
(x - 4)² + (y - 2) = 0
We can see that the center of the circle is now at the point (4, 2). So the new coordinates of the center are (h, k) = (4, 2).
Since the equation (x-4)² + (y-2) = 0 has no radius term (r²), it represents a degenerate circle where the radius is 0. In other words, it is not a circle, but a single point, i.e., the center of the circle.
Now, let's analyze the given options for the dimensions of A and B:
1) A = [LT], B = [LT]
2) A = [LT], B =[['T]
3) A = [2¹T¹,8-[LT]
4) A=[LT].B=[LT]
None of the given options seems to accurately represent the dimensions for the modified equation. As mentioned, the modified equation represents a degenerate circle with the center at (4, 2) and a radius of 0. So, the dimensions of A and B cannot be accurately determined based on the provided options.
It's possible that there might be a typo or misunderstanding in the options provided. If you have any additional context or clarification, please provide it so that I can offer a more precise explanation.
Explanation: