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

Answer:

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)#

#sqrt##17#.....................(Diameter)

so radius = 1/2 * diameter
= #sqrt##17#/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# = #(sqrt#17#/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

Answer:

#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#