Question

1. Identify the centers and radii of the following circles.

a. (x+25)^2+y^2=1
b. x^2+2x+y^2−8y=8
c. x^2−20x+y^2−10y+25=0
d. x^2+y^2=19
e. x^2+x+y^2+y=−1/4

Answer 1

a)
`(-25,0)\, ,r=1`

b)
`(-1,4)\, ,r=3`

c)
`(10,5)\, ,r=10`

d)
`(0,0)\, ,r=√(19)`

e)
`(-(1)/(2),-(1)/(2))\, ,r = (1)/(2)`

Explanation:

the general equation of a circle is can be represented in two forms

• 
  `(x-a)^2+(y-b)^2=r^2`

here,

the centre is:
`(a,b)`

radius is:
`r`

• 
  `x^(2)+y^(2)+2gx+2fy+c=0`

here,

the centre is:
`(-g,-f)`

radius is:
`√(g^2 + f^2 -c)`

Given these two forms we can solve the questions:

a)
`(x+25)^2+y^2=1`

this resembles the first equation of the circle hence we can compare them.

for center: we can see that +25 is the place of '-a', there is nothing in the place of b.

`-a = 25` solve for a

`a = -25`

similarly,

`b = 0`

the center is
`(a,b) = (-25,0)`

for radius:

`r^2 = 1`

`r = √(1)`

`r = 1`, we'll not take '-1' since that negative numbers don't apply to lengths.

b)
`x^2+2x+y^2-8y=8`

we can clearly see that this resembles the second general equation. Hence by comparison we can find +2x is in the place of +2gx, and solve for g.

`+2gx=+2x`

`g=1`

similarly

`+2fy=-8y`

`f=-4`

the centre of the circle is denoted by
`(-g,-f) = (-1,4)`

for radius:

`r = √(g^2 + f^2 -c)`

side note: the value of 'c' can be found if we simply move 8 from the right hand side of the equation to the left.
`x^2+2x+y^2-8y-8=0`

`r = √(1^2 + (-4)^2 -(-8))`

`r = √(9)`

`r = 3`

c)
`x^2-20x+y^2-10y+25=0`

we'll do these in the same manner as (b)

`+2gx=-20x`

`g=-10`

`+2fy=-10y`

`f=-5`

the centre of the circle is denoted by
`(-g,-f) = (10,5)`

for radius:

`r = √(g^2 + f^2 -c)`

`r = √((-10)^2 + (-5)^2 -25)`

`r = √((-10)^2 + (-5)^2 -25)`

`r = 10`

d)
`x^2+y^2=19`

this is easy in the sense that both general equations apply here. We can either say since there is no 2gx and 2fy term, both g and f are zeros hence the centre is (0,0)

or we can also say there's no 'a' and 'b', hence the centre is (0,0).

the centre is:
`(a,b) = (-g,-f) = (0,0)`

the radius can also be found through any of the two equations, but to keep it short we'll just use:

`r^2 = 19`

`r = \sqrt(19)`

e)
`x^2+x+y^2+y=-1/4`

this is one is again similar to the second equation.

for centre:

`+2gx = x`

`g = (1)/(2)`

`+2fy = y`

`f = (1)/(2)`

the centre of the circle is denoted by
`(-g,-f) = (-(1)/(2),-(1)/(2))`

for radius:

`r = √(g^2 + f^2 -c)`

`r = \sqrt{\left((1)/(2)\right)^2 + \left((1)/(2)\right)^2 -\left((1)/(4)\right)}`

`r = \sqrt{(1)/(4)\right}`

`r = (1)/(2)`