Question

x²+y²-4x+6y+12=0 then max value of |x-2|+|y+3| is​

Answer 1

Solution :

x² + y² - 4x + 6y + 12 = 0

=> ( x² - 4x + 4 ) + ( y² + 6y + 8 ) = 0

=> ( x² - 2(2x)(1) + (2)^2 ) + ( y² + 6y + 9 ) - 1 = 0

=> ( x - 2)² + ( y + 3)² = 1

Hence, the required maximum value is 1 .

This is the required answer .

_________________________

Additional Information :

(a + b)² = a² + 2ab + b²

(a + b)² = (a - b)² + 4ab

(a - b)² = a² - 2ab + b²

(a - b)² = (a + b)² - 4ab

a² + b² = (a + b)² - 2ab

a² + b² = (a - b)² + 2ab

2 (a² + b²) = (a + b)² + (a - b)²

4ab = (a + b)² - (a - b)²

ab = {(a + b)/2}² - {(a-b)/2}²

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

(a + b)³ = a³ + 3a²b + 3ab² b³

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)( a² - ab + b² )

a³ + b³ = (a + b)³ - 3ab( a + b)

a³ - b³ = (a - b)( a² + ab + b²)

a³ - b³ = (a - b)³ + 3ab ( a - b )

_________________________