The points $(0,4)$ and $(1,3)$ lie on a circle whose center is on the $x$-axis. What is the radius of the circle?

1 unit

Step-by-step explanation:

x² + (y-k)² = r²

0² + (4-k)² = r²

16 - 8k + k² = r²

1² + (3-k)² = r²

1 + 9 - 6k + k² = r²

r² - k² = 10 - 6k

16 - 8k = 10 - 6k

2k = 6

k = 3

r² - 3² = 10 - 6(3)

r² = 10 - 18 + 9

r² = 1

r = 1

Step-by-step explanation:

let the centre be (x,0) ...(∵ it lies on x-axis)

radius=√((x-0)²+(0-4)²)=√(x²+16)

also radius =√((x-1)²+(0-3)²)=√((x-1)²+9)

∴ √(x²+16)=√((x-1)²+9)

x²+16=(x-1)²+9

x²-(x-1)²=9-16

x²-(x²-2x+1)=-7

x²-x²+2x-1=-7

2x=-7+1=-6

x=-6/2=-3

so centre=(-3,0)

and radius=√(x²+16)=√((-3)²+16)=√(9+16)=√25=5

step-by-step explanation:

try remembering the steps

answer 1 : option "c"

answer 2 : option "d"

answer 3 : option "b"

answer 4 : option "d"

answer 5 : option "b"

answer 6 : option "a"

answer 7 : option "c"

answer 8 : option "d"

answer 9 : option "c"

answer 10 : option "d"

answer 11 : option "c"

answer 12 option "b"

answer 13 option "c"

answer 14 option "c"

answer 15 option "d"

answer 16 option "a"

answer 17 option "b"

answer 18 option "c"

answer 19 option "a"

answer 20 option "a"

answer 21 option "c"

d