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?

2 Answers

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lemjay | High School Teacher | (Level 3) Senior Educator

Posted on

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.





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`



`16=25`    (False)

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

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oldnick | (Level 1) Valedictorian

Posted on

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


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