Question

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?

Answer 1

Answer:

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

Answer 2

Answer:

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