Question

Find the equation of a circle whose centre is (2,1)
radius is 3
(a) x2 + y2 + 4x - 2y + 4 = 0
(b) x² + y2 - 4x + 2y - 4= 0
(c) x2 + y2 + 4x + 2y - 4 = 0
(d) x2 + y2 + 2x - 4y - 4 = 0​

Answer 1

Answer:

x² + y² - 4x -2y - 4 = 0 , none of these

Step-by-step explanation:

To find ---> Equation of circle whose centre is ( 2 , 1 ) and radius is 3

Solution ---> If coordinates of centre of circle is ( h , k ) and radius is r then equation of circle is

( x - h )² + ( y - k )² = r² ......................... (1)

It is given in question that

Ciordinates of centre = ( 2 , 1 )

radius = 3

So h = 2 , k = 1 , r = 3

Putting these values in equation (1)

=> ( x - 2 )² + ( y - 1 )² = 3²

We have an identity

(a - b )² = a² + b² - 2ab , applying it here

=> x² + 4 - 2 ( 2 ) x + y² + 1 - 2 (1 ) (y ) = 9

=> x² + 4 - 4 x + y² + 1 - 2y = 9

=> x² + y² - 4x - 2y + 5 - 9 = 0

=> x² + y² - 4x - 2y - 4 = 0

Additional information --->

General equation of circle

x² + y² + 2gx + 2fy + c = 0

Then

Coordinates of centre = (-g , - f )

Radius of circle = √(g² + f² - c )

Answer 2

Answer:

Mark me as brilliant ............,.