2 x322y28 find center radius and interceptsSolution2 x322y28

2 (x-3)^2+2y^2=8
find center, radius, and intercepts

2 (x-3)^2+2y^2=8

Thers is common term 2 with (x-3)^2 and y^2 which can be removed

Divide by 2 whole equation: (x-3)^2 + y^2 = 4

This equation is similar to standard equation o f circle with (h,k) centre and \'r\' as radius.

(x-h)^2 + (y-k)^2 = r^2

Comapring with stnd equation : (h, k) = ( 3, 0)

radius , r = sqrt4 =2

Intercepts: X intercept , plug y=0 : (x-3)^2 = 4----> x = 3+ / - 2 = 5 and 1

( 1, 0) and ( 5,0)

Y intercept , plug x=0 : y^2 =4 : y +/-2

( -2, 0) and ( 2, 0)