Question

The end points of a diameter of a circle are (1, -1)
and (3, 5). Find the equation of the circle. Also
find its centre and radius.​

Answer 1

Answer:

x^2 + y^2 - 4x - 4y - 2 = 0

Step-by-step explanation:

Equation of the circle ( when ends points of the diameter are given ) is given by

• ( x - a )( x - c ) + ( y - b )( y - d ) = 0 , coordinates of the ends points are ( a, b ) and ( c, d ).

Here,

Coordinates are ( 1 , - 1 ) and ( 3 , 5 ).

Thus,

= > Equation of the circle

= > ( x - 1 )( x - 3 ) + { y - ( - 1 ) }( y - 5 ) = 0

= > x^2 - 3x - x + 3 + ( y + 1 )( y - 5 ) = 0

= > x^2 - 4x + 3 + y^2 - 5y + y - 5 = 0

= > x^2 - 4x + 3 + y^2 - 4y - 5 = 0

= > x^2 + y^2 - 4x - 4y - 2 = 0

Coordinates of its centre = coordinates of mid point of diameter = { ( 3 + 1 ) / 2 , ( 5 - 1 ) / 2 } = ( 2 , 2 )

Radius = Distance between centre and a point of circle = √{( 2 - 1 )^2 + ( 2 + 1 )^2 } = √( 1 + 9 ) = √10 unit.