Please helppppp

Question

find the standard equation of the circle through (0,1) , (2,3) and (1,1+√3)

Answer 1

There are at least 3 ways to solve this problem; none is other than messy.

First.

You can plug the 3 co-ordinates into the equation (x-h)² + (y-k)² = r^{2 to get a set of 3 simultaneous equations and solve for the 3 parameters. The problem with this method is that the equations are quadratics...so not so easy to solve.}

^Second

^{Compute the equation of the straight lines through any 2 pairs of co-ordinates.}

^{Compute the perpendicular bisectors of these 2 lines.}

^{The intersection of those bisectors will give you the center of the circle (h and k in the equation above).}

^{The distance formula from the center to any of the 3 points will give you the radius (r in the equation above).}

Third:

The general equation of the circle is x² + y² + ax +by + c = 0

Substitution of the co-ordinates into this equation yields 3 LINEAR equations which can be solved

by determinants (or any other convenient way) to get the parameters a, b and c.

The parameters h, k and r can then be obtained by completing the square on the x terms and then on the y terms.

All of these methods, as stated above, are messy, but will get you the answer.

Answer 2

What is noted in the first answer is correct. If you designate the center of the circle as (x_o,y_o) and then use 2 of the points and use the distance formula which is r (radius) of the circle, you get one linear equation in (x_o,y_o)

Then use the third point in conjunction with either of the other 2 points, you get another linear equation in (x_o,y_o)

So, use these 2 equations to determine (x_o,y_o) which turns out to be (2,1)

Then substitute back in the distance formula to get the radius which is 2.