WRITE AN EQUATION OF THE CIRCLE WITH CENTER AND RADIUS

Find the standard form equation for the circle.

(i)  Center (1, 2) and radius 5.

(ii)  Center (-3, 2) and radius 1.

(iii)  Center (-1, -4) and radius 3.

(iv)  Center (0, 0) and radius √3.

Solution :

(i)  Center (1, 2) and radius 5.

Equation of the circle :

(x-h)² + (y-k)²  =  r²

(x-1)² + (y-2)²  =  5²

(x-1)² + (y-2)²  =  25

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

x²-2x+1+y²-4y+4  =  25

x²+y²-2x-4y+5-25  =  0

x²+y²-2x-4y-20  =  0

So, the required equation of the circle is

x²+y²-2x-4y-20  =  0

(ii)  Center (-3, 2) and radius 1.

Solution :

Here h = -3 and k = 2

Equation of the circle :

(x-h)² + (y-k)²  =  r²

(x+3)² + (y-2)²  =  1²

(x+3)²  =  x²+6x+9 and (y-2)² =  y²-4y+4

x²+6x+9+y²-4y+4  =  1

x²+y²+6x-4y+9+4-1  =  0

x²+y²+6x-4y+12  =  0

So, the required equation of the circle is

x²+y²+6x-4y+12 =  0

(iii)  Center (-1, -4) and radius 3.

Solution :

(x-h)² + (y-k)²  =  r²

(x+1)² + (y+4)²  =  3²

(x+1)²  =  x²+2x+1 and (y+4)² =  y²+8y+16

x²+2x+1+y²+8y+16  =  9

x²+y²+2x+8y+17-9  =  0

x²+y²+2x+8y+8  =  0

So, the required equation of the circle is

x²+y²+2x+8y+8  =  0

(iv)  Center (0, 0) and radius √3.

Solution :

(x-h)² + (y-k)²  =  r²

(x-0)² + (y-0)²  =  √3²

x² + y²  =  3

So, the required equation of the circle is

x² + y²  =  3

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