What is the radius of the circle x^2+y^2+21=10x

The radius of the given equation is 2 .

Step-by-step explanation:

The equation of a circle whose center (h, k) and radius is r,

(x - h)² + (y - k)² = r² (i)

Here given that-

x² + y² + 21 = 10 x

x² - 10 x + y² + 21 = 0

( To making perfect square, add and subtract 25 ) -

x² - 10 x + 25 - 25 + y² + 21 = 0

(x² - 10 x + 25) + y² -4 =0

[As we know that -

(x - 5)² = x² - 10 x + 25]

(x - 5 )² + y² = 4

(x - 5)² + (y - 0)² = 4

(x - 5 )² + (y- 0)² = 2² (ii)

On comparing equation (i) and(ii)-

r = 2

Hence the radius of the given equation is 2 and center is ( 5, 0).

c.

step-by-step explanation:

substitution for (0,1)

y=x^2+4x+1

1=0^2+4(x)0+1

1=0+4(x)0+1

1=4(x)0+1

1=0+1

1=1

substitution for (-3,-2)

y=x+1

-2=-3+1

-2=-2