The equation of a circle is given , where a is the radius. If the equation is modified to change the origin other than (0, 0), then find out the correct dimensions of A and B in a new equation : The dimensions of t is given as []. [2023]

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Q.
The equation of a circle is given by $x^{2} + y^{2} = a^{2}$ , where $a$ is the radius. If the equation is modified to change the origin other than (0, 0), then find out the correct dimensions of A and B in a new equation : ${(x - A t)}^{2} + {(y - \frac{t}{B})}^{2} = a^{2}$

The dimensions of t is given as [ $T^{- 1}$ ]. [2023]

1 $A = [L^{- 1} T], B = [L^{- 1} T]$

2 $A = [L T], B = [L^{- 1} T^{- 1}]$

3 $A = [L^{- 1} T^{- 1}], B = [L^{- 1}]$

4 $A = [L^{- 1} T^{- 1}], B = [L T]$

Ans.
(2)

${(x - A t)}^{2} + {(y - \frac{t}{B})}^{2} = a^{2}$

$[A t] = A \times \frac{1}{T} = L$

$∴ [A] = T^{1} L^{1}$

$\frac{t}{B}$ is in meters

$∴ \frac{1}{T [B]} = L \Rightarrow [B] = T^{- 1} L^{- 1}$

Solution

1 $A = [L^{- 1} T], B = [L^{- 1} T]$

2 $A = [L T], B = [L^{- 1} T^{- 1}]$

3 $A = [L^{- 1} T^{- 1}], B = [L^{- 1}]$

4 $A = [L^{- 1} T^{- 1}], B = [L T]$

Ans.
(2)

${(x - A t)}^{2} + {(y - \frac{t}{B})}^{2} = a^{2}$

$[A t] = A \times \frac{1}{T} = L$

$∴ [A] = T^{1} L^{1}$

$\frac{t}{B}$ is in meters

$∴ \frac{1}{T [B]} = L \Rightarrow [B] = T^{- 1} L^{- 1}$

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Solution

Q. The equation of a circle is given by x2+y2=a2, where a is the radius. If the equation is modified to change the origin other than (0, 0), then find out the correct dimensions of A and B in a new equation : (x–At)2+(y–tB)2=a2 The dimensions of t is given as [T–1]. [2023]

1 A=[L–1T], B=[L–1T]

2 A=[LT], B=[L–1T–1]

3 A=[L–1T–1], B=[L–1]

4 A=[L–1T–1], B=[LT]

Ans. (2) (x–At)2+(y–tB)2=a2 [At]=A×1T=L ∴ [A]=T1L1 tB is in meters ∴ 1T[B]=L ⇒ [B]=T–1L–1

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1 $A = [L^{- 1} T], B = [L^{- 1} T]$

2 $A = [L T], B = [L^{- 1} T^{- 1}]$

3 $A = [L^{- 1} T^{- 1}], B = [L^{- 1}]$

4 $A = [L^{- 1} T^{- 1}], B = [L T]$

Ans.
(2)

${(x - A t)}^{2} + {(y - \frac{t}{B})}^{2} = a^{2}$

$[A t] = A \times \frac{1}{T} = L$

$∴ [A] = T^{1} L^{1}$

$\frac{t}{B}$ is in meters

$∴ \frac{1}{T [B]} = L \Rightarrow [B] = T^{- 1} L^{- 1}$