Find the value of x,y if the distances of the point (x,y) from (-3,0) as well as from (3,0) are 4

Find the value of x,y if the distances of the point (x,y) from (-3,0)

1.	Find the value of x,y if the distances of the point (x,y) from (-3,0) as well as from (3,0) are 4
Answer» We have P(x, y), Q(-3, 0) and R(3, 0){tex}P Q = \\sqrt { ( x + 3 ) ^ { 2 } + ( y - 0 ) ^ { 2 } }{/tex}{tex}\\Rightarrow \\quad 4 = \\sqrt { x ^ { 2 } + 9 + 6 x + y ^ { 2 } }{/tex}Squaring both sides{tex}\\Rightarrow \\quad ( 4 ) ^ { 2 } = \\left( \\sqrt { x ^ { 2 } + 9 + 6 x + y ^ { 2 } } \\right) ^ { 2 }{/tex}{tex}\\Rightarrow{/tex}\xa016 = x2 + 9 + 6x + y2{tex}\\Rightarrow{/tex}x2 + y2 = 16 - 9 - 6x{tex}\\Rightarrow{/tex}x2 + y2 = 7 - 6x ...(i){tex}P R = \\sqrt { ( x - 3 ) ^ { 2 } + ( y - 0 ) ^ { 2 } }{/tex}\xa0{tex}\\Rightarrow{/tex}{tex}4 = \\sqrt { x ^ { 2 } + 9 - 6 x + y ^ { 2 } }{/tex}Squaring both sides{tex}( 4 ) ^ { 2 } = \\left( \\sqrt { x ^ { 2 } + 9 - 6 x + y ^ { 2 } } \\right) ^ { 2 }{/tex}{tex}\\Rightarrow{/tex}16 = x2 + 9 - 6x + y2{tex}\\Rightarrow{/tex}x2 + y2 - 16 - 9 + 6x ...(ii)Equating (i) and (ii)7 - 6x = 7 + 6x{tex}\\Rightarrow{/tex}7 - 7 = 6x + 6x{tex}\\Rightarrow{/tex}0 = 12x{tex}\\Rightarrow{/tex}x = 0Substituting the values of x = 0 in (ii)x2 + y2 = 7 + 6x0 + y2 = 7 + 6\xa0{tex}\\times{/tex}0y2\xa0= 7y =\xa0{tex}\\pm \\sqrt { 7 }{/tex}