Question

What is the center and radius of the circle with the given equation? D8" id="TexFormula1" title="x^2+y^2+2x+8y=8" alt="x^2+y^2+2x+8y=8" align="absmiddle" class="latex-formula">

Answer 1

Answer:

(- 1, - 4 ) and r = 5

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

To obtain this form from the given equation use the method of completing the square.

Given

x² + y² + 2x + 8y = 8 ( rearrange by collecting x and y terms )

x² + 2x + y² + 8y = 8

To complete the square

add ( half the coefficients of the x/ y terms )² to both sides

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

(x + 1)² + (y + 4)² = 25 ← in standard form

with centre (h, k ) = (- 1, - 4 ) and r = $\sqrt{25}$ = 5