The given equation of the circle is
` 2x ^(2) + 2 y ^(2) - 6x + 8y + 1 = 0 `
` rArr x ^(2) + y ^(2) - 3x + 4y + (1)/(2) = 0 rArr x ^(2) + y ^(2) + 2gx + 2f y + c = 0`, where ` g = (-3)/(2), f = 2 and c = (1)/(2)` .
Centre of this circle `= (-g, - f ) = ((3)/(2), - 2 )`
Radius of this circle `= sqrt (g ^(2) + f ^(2) - c )`
` " " = sqrt ((9)/(4) + 4 - (1)/(2)) = (sqrt ( 23 ))/( 2)`
` therefore ` the centre of the required circle ` = ((3)/(2), - 2)`.
Let ` r_1 ` be the radius of the required circle. Then,
`pi r _1 ^(2) = 2 { pi xx ((sqrt ( 23))/(2)) ^(2) } rArr r_1 ^(2) = ( 23)/(2)`
Hence , the equation of the required circle is
` (x - (3)/ ( 2) ) ^(2) + (y + 2 ) ^(2) = ( 23)/(2)`
` rArr x ^(2) + y ^(2) - 3x + 4y - ( 21)/(4) = 0 `
` rArr 4x ^(2) + 4y ^(2) - 12 x + 16 y - 21 = 0 `
