find the centre and radios of circle2(x+1)^2+ 2y^2=25
Answers
Answered by
3
rowseNotessearch
HOMEWORK HELP > MATH
Find the centre and radius of the circle:x^2 + y^2 + 8x + 6y= 0
print Print
document PDF list Cite
EXPERT ANSWERS
HALA718 | CERTIFIED EDUCATOR
x^2 + y^2 + 8x + 6y= 0
We need to re-write the equation in the following format:
( x- a)^2 + ( y-b)^2 = r^2 such that:
(a,b) is the center of the circle and r is the radius:
Let us calculate:
We will re-arrange terms:
x^2 + 8x + y^2 + 6y = 0
We will complete the squares:
==> ( x+4)^2 - 16 + ( y+3)^2 - 9 = 0
==> ( x+4)^2 + ( y+3)^2 - 25 = 0
We will move 25 to the right side:
==> ( x+ 4)^2 + ( y+3)^2 = 25
==> ( x+ 4)^2 + ( y+3)^2 = 5^2
Now the equation is in the standard form:
Then the center of the circle is ( -4, -3) and the radius = 5
HOMEWORK HELP > MATH
Find the centre and radius of the circle:x^2 + y^2 + 8x + 6y= 0
print Print
document PDF list Cite
EXPERT ANSWERS
HALA718 | CERTIFIED EDUCATOR
x^2 + y^2 + 8x + 6y= 0
We need to re-write the equation in the following format:
( x- a)^2 + ( y-b)^2 = r^2 such that:
(a,b) is the center of the circle and r is the radius:
Let us calculate:
We will re-arrange terms:
x^2 + 8x + y^2 + 6y = 0
We will complete the squares:
==> ( x+4)^2 - 16 + ( y+3)^2 - 9 = 0
==> ( x+4)^2 + ( y+3)^2 - 25 = 0
We will move 25 to the right side:
==> ( x+ 4)^2 + ( y+3)^2 = 25
==> ( x+ 4)^2 + ( y+3)^2 = 5^2
Now the equation is in the standard form:
Then the center of the circle is ( -4, -3) and the radius = 5
dtjk:
your answer is really not clear
Similar questions