maths help(circles)

Question

What is the equation of a circle, when the radius is 5 and the points pass through (3, 1) and (-2, -4)?

Answer 1

Shivam,

    What math class are you taking? You come up with an interesting variety of questions.

    (x-h)^2  + (y-k)^2  = r^2  When a grapg goes through a point, the coordinate must satisfy the equation, so,

(3-h)^2 + (1-k)^2 = 25    and     (-2-h)^2 + (-4-k)^2 = 25 

9-6h+h^2 + 1-2k+k^2 = 25        4+4h+h^2 + 16+8k+k^2 = 25

                                                       9-6h+h^2 + 1-2k+k^2     =25

subtract these to get                      -5+10h       +15 +10k       = 0

                                                                      10h +10k            = -10

                                                                               h+k              = -1  or h=-k-1

 

substitute this back into one of the equations to get

(3 - (-k-1))^2 + (1-k)^2 = 25 

(4+k)^2 + (1-k)^2 = 25 

16+8k+k^2 + 1-2k+k^2 = 25 

2k^2 + 6k -8 = 0

k^2 +3k -4 = 0

(k+4)(k-1)=0

k=-4 or k=1

 

2 possible centers.....(h,-4) or (h,1)

plug into h=-k-1 to find the corresponding h values and get

(3,-4) or (-2, 1) as possible centers....

(x-3)^2 + (y+4)^2 =25    and (x+2)^2 + (y-1)^2 = 25 are your 2 possible answers.

you can easily check to see that your original points satisfy both of these equations.

Answer 2

the general equation of a circle as you know is (x-h)² + (y-k)² = r²

 

the distance between the two points given is 5√2

 

the diameter is 10

 

therefore the line segment between the points (a chord) is not the diameter.

 

The circle satisfying these conditions is not unique. If you draw a perpendicular through the midpoint of the chord, it will contain two possible center points each equidistant from the chord.