Equation of a circle

[_JJ]

find the equation of the circle that passes through the points A,B,C : A(-2,2) B(2,4) C(2,-5)
I have done this question many times now but the answer I get is different to the answer given in the back of the book. The answer given is: 2x^2 +2y^2 -7x+2y-34=0

 

[Harry_the_cat]

You can check the answer by checking each point satisfies the equation. They do, so the answer is correct.

Show us your work and we can help see where you are going wrong.

 

[pka]

find the equation of the circle that passes through the points A,B,C : A(-2,2) B(2,4) C(2,-5)
I have done this question many times now but the answer I get is different to the answer given in the back of the book. The answer given is: 2x^2 +2y^2 -7x+2y-34=0

Look at this link Compare your result & the book's result to that link.
If any are different, the check to see if you have copied the coordinates correctly.

 

[MarkFL]

find the equation of the circle that passes through the points A,B,C : A(-2,2) B(2,4) C(2,-5)
I have done this question many times now but the answer I get is different to the answer given in the back of the book. The answer given is: 2x^2 +2y^2 -7x+2y-34=0

I'd write:

[MATH](2+h)^2+(2-k)^2=r^2[/MATH]
[MATH](2-h)^2+(4-k)^2=r^2[/MATH]
[MATH](2-h)^2+(5+k)^2=r^2[/MATH]
Solving this system, we find:

[MATH](h,k,r^2)=\left(\frac{7}{4},-\frac{1}{2},\frac{325}{16}\right)[/MATH]
And so the equation of the circle is:

[MATH]\left(x-\frac{7}{4}\right)^2+\left(y+\frac{1}{2}\right)^2=\frac{325}{16}[/MATH]
Or:

[MATH]2x^2+2y^2-7x+2y-34=0[/MATH]

 

[Steven G]

Can you please show us your work so that we can find your mistake, as Harry_the_cat has mentioned. How else can we help you?! We can solve the problem for you but then you do not see where you made your mistake. Seeing your mistake hopefully would prevent you from making the same mistake.