Find the equation of the circle with diameter AB where A and B are the points (-1,2) and (3,3) respectively?

2 Answers
Sep 15, 2015

The answer could be find out by knowing some formulas of coordinate geometry.
1) standard eqn.= (x−a)^2 + (y−b)^2 = r^2 (for this question i,e, radius)
2) midpoint= (x_1+x_2)/2,( y_1+y_2)/2
3) distance= sqrt(x_2-x_1)^2 + (y_2-y_1)^2

Explanation:

we are given , A(-1,2) & B(3,3)
here x_1= -1
y_1= 2
x_2= 3
y_2= 3

Therefore, AB= sqrt(x_2-x_1)^2 + (y_2-y_1)^2

= sqrt(3 -(-1))^2 + (3-2)^2

sqrt(4)^2 + (1)^2

sqrt(16) + (1)

sqrt17.....................(Diameter)

so radius = 1/2 * diameter
= sqrt17/2

now the radius can also be got by calculating the midpoint
so the next step is:
wkt,midpoint is given by =(x_1+x_2)/2,( y_1+y_2)/2

then,we get (-1+3)/2 , (2+3)/2

= (2/2) , (5/2)

= (1 , 2.5)
Hence we get the values of a and b respectively.
Putting the values we get,
(x−a)^2 + (y−b)^2 = r^2

= (x−1)^2 + (y−2.5)^2 = (sqrt17/2)^2
= (x−1)^2 + (y−2.5)^2 = (17/4)
= (x−1)^2 + (y−2.5)^2 = 4.25
= (x−1)^2 + (y−2.5)^2 = 4.25........................(answer)

Sep 17, 2015

x^2+y^2-2x-5y+3=0

Explanation:

Firstly, we can find the centre of the circle by finding the midpoint of AB. Since it the midpoint or the centre (h,k) cut AB with equal ratio;

Centre (h,k) = ((-1+3)/2,(2+3)/2)

(h,k)=(1,5/2)

Then we can find the radius , r of the circle by using the equation;

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

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

Substitute the coordinate (h,k)=(1,5/2) and any of A or B coordinates into equation. In this calculation I choose B.

r^=sqrt((3-1)^2+(3-5/2)^2

r=sqrt(17/4)=sqrt17/2

To find c, we can use the equation c=h^2+k^2-r^2 and substitute (h,k)=(1,5/2) and r=sqrt17/2 into it.

c=(1)^2+(5/2)^2-(sqrt17/2)^2

c=3

Then we know that the equation of circle is known as;

x^2+y^2-2hx-2ky+c=0

Substitute only (h,k)=(1,5/2) and c=3 and we get;

x^2+y^2-2(1)x-2(5/2)y+3=0

x^2+y^2-2x-5y+3=0