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If the equation of a circle is ax2 + (2a - 3)y2 - 4x - 1 = 0, then its centre is:
1. (1, 1)
2. (2, 0)
3. \(\left( {\frac{2}{3},0} \right)\)
4. \(\left( { - \frac{2}{3},0} \right)\)

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Correct Answer - Option 3 : \(\left( {\frac{2}{3},0} \right)\)

Concept:

General form of the equation of a circle, x2 + y+ 2gx + 2fy + c = 0     ------(1)

  • The centre is (-g, -f) or \(\left( {\frac{{ - \;{\rm{coefficient\;of\;x}}}}{2},\;\frac{{ - \;{\rm{coefficient\;of\;y}}\;}}{2}} \right)\) and radius =\(\sqrt {{g^2} + {f^2} - c} \), where g, f and c are three constants

Standard Form of the equation of a circle, (x – h)2 + (y – k)2 = r2

  • Where the centre (h, k) and the radius r.

Calculation:

Given equation of circle is ax2 + (2a - 3)y2 - 4x - 1 = 0    ------(a)

from circle a = 2a - 3 ⇒ a = 3

put in equation (a) we get

⇒ 3x2 + 3y2 - 4x - 1 = 0

⇒ x2 + y2 - (4/3)x - 1/3 = 0 

on comparing with equation (1) we get,

⇒ 2g = -(4/3) ⇒ g = -(2/3)

and 2f = 0 ⇒ f = 0

Thus centre (-g, -f) = (2/3, 0)

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