After an unsuccessful search for the answer to my math problem I came here to ask my question. I am trying to find the equation of circle but having a difficulty.
The Problem is as follows:
Find the Equation of Circle which passes through the points A(1, -2), B(4,3) and has its center on the line 3x+4y-7=0.
I want to solve it by two methods. I was able to solve it by the first method but got different values in the 2nd method.
My Approach using the first method:
let the required equation of circle be $$(x-h)^2+(y-k)^2=r^2$$
I draw this as a rough diagram so you guys can understand better
Here in the above diagram: $$|AO| = |BO|$$
$$=> \sqrt{((-g)-1)^2 + ((-f)+2)^2 } = \sqrt{((-g)-4)^2 + ((-f)-3)^2 }$$
let (h,k) = (-g,-f) then
$$=> \sqrt{(h-1)^2 + (k+2)^2 } = \sqrt{(h-4)^2 + (k-3)^2 }$$
Squaring both sides of the above equations
$$=> (\sqrt{(h-1)^2 + (k+2)^2 })^2 = (\sqrt{(h-4)^2 + (k-3)^2 })^2$$
Therefore $$=>h^2+1-2h+k^2+4+4k = h^2+16-8h+k^2+9-6k$$
$$=> -6h-10k+20=0$$
$$=> 3h+5k-10=0——(i)$$
Given Equation of Line is $$3x+4y-7=0——(ii)$$
$$(ii)=> 3h+4k-7=0 $$ (As the line passes through the center of the circle) and
C(-g,-f) = C(h,k)
$$(i)=> 3h+5k=10$$
$$(ii)=> 3h+4k=7$$
Now Solving (i) and (ii) simultaneously to find the coordinates of the center of the circle
Subtract (ii) from (i)
Therefore $$3h+5k=10$$
$$- (3h+4k)=-(7)$$
$$———–$$
$$k = 3$$
Put k=3 in equation (i)
$$(i)=> 3h+5(3)=10$$
$$=> h = -5/3$$
Now finding the radius of the Circle:
$$|AO| = r = \sqrt{(-5/3 – 4)^2 + (3-3)^2}$$
$$=> r = -5/3 – 4$$
$$=> r = -17/3$$
Now we can find the circle by using standard form
that is $$(x-h)^2(y-k)^2 = r^2$$
$$=> (x+5/3)^2 + (y-3)^2 = (17/3)^2$$
$$=> x^2+25/9+10x/3+y^2+9-6y = 289/9$$
$$=> x^2+y^2+10x/3-6y = 289/9-25/9-9$$
$$=> x^2+y^2+10x/3-6y = 183/9$$
$$=> x^2+y^2+10x/3-6y-183/9= 0 $$ (Multiply both sides by 3)
$$=> 3x^2+3y^2+10x-18y-183/3= 0 $$
Required Equation of Circle is $ 3x^2+3y^2-10x-18y-61= 0 $
My Approach using the 2nd method:
In this method I used general form of equation of the circle to derive the two equations w.r.t 2 points on the circle A and B.
General Form of equation of Circle: $$x^2+y^2+2gx+2fy+c=0$$
for point A(1,-2) on the circle, general form is reduced to: $$1+4+2g-4f+c=0$$
$$=> 2g-4f+c = -5 —–(i) $$
for point B(4,3) on the circle, general form is reduced to: $$16+9+8g+6f+c=0$$
$$=> 8g+6f+c = -25—–(ii)$$
equation of line = $3x+4y-7=0 => -3g-4f=7 => 3g+4f=-7 —–(iii)$
(i), (ii) and(iii) forms a system of linear equations in 3 variables (g,f and c), we need to find those three variable for that reason we write the above the equation in matrix form
$$ A= \begin{bmatrix}2&-4&1\\8&6&1\\3&4&0\end{bmatrix} X= \begin{bmatrix}g\\f\\c\end{bmatrix} B= \begin{bmatrix}-5\\-25\\-7\end{bmatrix} $$
Writing in Augmented form
$$[A|b] =
\left[\begin{array}{rrr|r}
2 & -4 & 1 & -5 \\
8 & 6 & 1 & -25 \\
3 & 4 & 0 & -7
\end{array}\right]
$$
Now we have to write the above [A|b] in echelon form to find g, f and c and then put them in the general form of the circle in order to get the equation of the circle…This is the step where it give me the wrong answer can you guys check what am I doing wrong?
Edit: ** Credit goes to David in the answer below. There were two minor mistakes pointed out by @David as I wrote 7 instead of -7 in augmented matrix and in **$$(x-(-5/3))^2+(y-3)^2$$
I wrote $$(x-5/3)$$ instead of $$(x+5/3)$$
Best Answer
You made a mistake in your first attempt.
You determined that the center of the circle is at $(h,k) = \left(-\frac53, 3\right)$ and the radius is $\frac{17}3$. Then you wrote the equation
$$ \left(x - \frac53\right) + (y−3)^2 = \left(\frac{17}3\right)^2. $$
But $x - \left(- \frac53\right)^2 = x + \frac53,$ so the equation for the circle is actually $$ \left(x + \frac53\right)^2 + (y−3)^2 = \left(\frac{17}3\right)^2. $$
In your second attempt, what you have shown is correct almost to the end. Indeed,
$$ A^{-1}b = \begin{pmatrix} 2& -4 & 1 \\ 8 & 6 & 1 \\ 3 & 4 & 0 \end{pmatrix}^{-1} \begin{pmatrix} -5 \\ -25 \\ -7\end{pmatrix} = \begin{pmatrix} \frac53 \\ -3 \\ -\frac{61}3\end{pmatrix}, $$
which is the correct result. But for some reason, as noted in a comment you have $7$ instead of $-7$ when you built the augmented matrix. That's definitely going to cause a wrong result.