Find a positive value of \(\alpha\) if the coordinate of the centre of a circle x\(^2\) + y\(^2\) - 2\(\alpha\)x + 4y - \(\alpha\) = 0 is (\(\alpha\), -2) and the radius is 4 units.

Antoni19

5 years ago

r = √(x² + y² - c)
4 = √(α² + (-2)² - (-α))
4 = √(α² + α + 4)
Square both sides
16 = α² + α + 4
α² + α - 12 = 0
Factorize
(α + 4)(α - 3) = 0
α = 3

Chibou

4 years ago

I don't understand

Joshie777

12 years ago

We do need an explanation here plss

Meez fola bankz

9 years ago

a=3

chukwujekwuChristian

1 year ago

more explanation

olowolafe olumide michael

13 years ago

i think the answer is -1 but it is not a positive integer

Chascopaul

7 years ago

if x^2+y^2-2ax+4y-a=0
and center is (a,-2)
then using method of completing the square
that is by adding to both sides the square of the half of co officient of x and y. we have
x^2-2ax+{1/2*-2a}^2+y^2+4y+{1/2*4}^2=a+a^2+4
factorizing,
{x-a}^2+{y+2}^2=a+a^2+4
but radius =4
and equation of circle is (x-a)^2+(y-b)^2=r^2
comparing with the above we have
a+a^2+4=4^2
arranging,
a^2+a-12=0
a=3 or a=-4
so therefore the positive value of a=3
answer=c

RALPHAEL

2 months ago

dont understand

lildenex009

2 months ago

what of the explanation video

berber

11 years ago

X^2 y^2-2ax 4y-a=0

x^2-2ax y^2 4y=a

take half of coefficient of x and y,square and add to both sides,

x^2-2ax (-a)^2 y^2 4y (2)^2=a a^2 4.

(x-a)^2 (y 2)^2=a^2 a 4.

Compare with (x-a)^2 (y-b)^2=r^2.

a^2 a 4=r^2,

a^2 a 4=4^2,

a^2 a 4=16,

a^2 a 4-16=0,

a^2 a-12=0,

a^2 4a-3a-12=0,

a(a 4)-3(a 4)=0,

(a-3)(a 4)=0,

a-3=0 or a 4=0,

a=3 or -4

a= 3(positive value).