# Find the equation of a circle which center is at C(2, -4) and passes through the point P(-1,0). Is the point K(6,-8) on the circle?

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To determine the equation of circle use the formula:

`(x - h)^2 + (y-k)^2=r^2`

where (h,k) is the center, (x,y) are the points along the circle and r is the radius.

Since the radius of the circle is not given, substitute the center and the given point to determine the value of r^2.

`(-1-2)^2+(0-(-4))^2=r^2`

`(-3)^2+(4)^2=r^2`

`9+16=r^2`

`25=r^2`

Now that the value of r^2 is known, plug-in the value of h, k and r^2 to the formula.

`(x -2)^2+(y-(-4))^2=25`

`(x-2)^2+(y+4)^2= 25` **Hence, the equation of the given circle is `(x-2)^2+(y+4)^2= 25` .**

Next, to determine if the point (6,-8) is on the circle, substitute this point to the equation of the circle.

`(x-2)^2+(y+4)^2= 25`

`(6-2)^2+(-8+4)^2=25`

`4^2+4^2=25`

`16=25` (False)**Since the resulting condition is False, hence the point K(6,-8) does not lie along the circle.**

firast we have to find radius:

r =`sqrt[(2+1)^2 + 4^2 ` = 5

so the eqyurtaion is:

(x-2)`^2` + (y+4) `^2` = 25

x`^2` -4x +4 + y`^2` +8y +16 = 25

then:

x`^2` +y`^2` -4x +8y -9 = 0

is the equation we are searching for.

to chek point K lies on cirumference really mate, is enough to chek if the distance from the center C is just r = 5

d(K;C) = `sqrt[(6-2)^2 + (-8 +4) ^2` `=` `4` `sqrt(2)`

We find it doesn't lie on the cirucmfernece