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Find the equation of the circle passing through the points $P(5,7)$, $Q(6,6)$ and $R(2,-2)$.

My Attempt:

Let the equation of the circle be: $$x^2+y^2+2gx+2fy+c=0$$

The point $P(5,7)$ lies on the circle then, $$5^2+7^2+10g+14f+c=0$$ $$10g+14f+c=-74$$-----(1)

The point $Q(6,6)$ lies on the circle then, $$6^2+6^2+12g+12f+c=0$$ $$12g+12f+c=-72$$-------(2)

The point $R(2,-2)$ lies on the circle the, $$2^2+(-2)^2+4g-4f+c=0$$ $$4g-4f+c=0$$-----(3).

Now, please help me from here.

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  • $\begingroup$ You get a system of 3 equations and 3 unknowns that you can solve. $\endgroup$ – Markus Oct 1 '16 at 0:29
  • $\begingroup$ @ Markus, I get problem in solving those equations. Could not get the answer. $\endgroup$ – pi-π Oct 1 '16 at 0:37
  • $\begingroup$ Using LinAlg and a computer to row reduce, I get $g=-7/3,f=-10/3,c=-4$. $\endgroup$ – Bobson Dugnutt Oct 1 '16 at 0:45
  • $\begingroup$ "Markus, I get problem in solving those equations. Could not get the answer" What'd you try. Should be straightforward. subtract 2 from 1 and you get 2g - 2f = 2 so g - f = 1. Multiply by 4 and get 4g - 4f = 4 Plug that into 3 and you get 4+c = so c = - 4. So 10g + 14f = -70 and 12g +12f = -68. Keep going. $\endgroup$ – fleablood Oct 1 '16 at 1:08
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Another way to do this is to use the geometric fact that the perpendicular bisector of a chord passes through the center of a circle. The line through P(5,7) and Q(6,6)I has slope (7- 6)/(5- 6)= -1 and midpoint (11/2, 13/2). The perpendicular bisector is y= (x- 11/2)+ 13/2)= x+ 1. The line through Q(6, 6) and R(2,−2) has slope (6-(-2))/(6- 2)= 8/4= 2 and midpoint (4, 2). The perpendicular bisector is y= -(1/2)(x- 4)+ 2= -(1/2)x+ 4.

Those two lines intersect when y= x+ 1= -(1/2)x+ 4 so (3/2)x= 3, x= 2. Then y= 2+ 1= 3. The center of this circle is (2, 3) and the radius is $\sqrt{(2- 6)^2+ (6- 3)^2}= \sqrt{25}= 5$. The equation of that circle is $(x- 2)^2+ (y- 3)^2= 25$.

As a check $(5- 2)^2+ (7- 3)^2= 9+ 16= 25$, $(6- 2)^2+ (6- 3)^2= 16+ 9= 25$, and $(2- 2)^2+ (-2- 3)^2= 0+ 25= 25$.

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Subtract $2^{nd}$ equation from $3$ times the $3^{rd}$ to get

$$ c-12f=36 $$

Then subtract $2$ times the first from $5$ times the third to get:

$$ 3c-48f=148 $$

From the last two, subtract $3$ times the first from the second to get $f=-10/3$. Plugging back in one them gives $c=-4$. Now pluging in one of the first three, gives $g=-7/3$.

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Just a different approach: When you construct perpendicular bisectors of PQ and QR, then these lines intersect each other exactly in the center of the circle. So midpoint of PQ is $(5.5,6)$, slope of line PQ is $-1$ and perpendicular bisector is then given by: $y-6=1(x-5.5)$. Midpoint of QR is $(4,2)$, slope of line QR is $2$ and perpendicular bisector is given by $y-2=2(x-4)$. Finding the intersection of these lines is easy by solving $x-5.5+6=2(x-4)+2$. Once you have the coordinates of this intersection, you have the center of the circle and the $r^2$ (radius squared) can be found by Pythagorean theorem which in essence gives you the equation of the circle. Can you solve it from here?

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Another way could be successive elimination. Considering the three equations (please, notice that your third equation is wrong) $$10g+14f+c+74=0\tag 1$$ $$12g+12f+c+72=0\tag 2$$ $$4g-4f+c+8=0\tag 3$$ From $(1)$, eliminate $$c=-14 f-10 g-74\tag 4$$ Replace in $(2)$ and $(3)$ to get $$-2 f+2 g-2=0\tag 5$$ $$10 f+14 g+74=0 \tag 6$$ From $(5)$, eliminate $$f= g-1\tag 7$$ Replace in $(6)$ $$24 g+48=0\tag 8$$ So $g=-2$ and going backwards $f=-3$ and $c=-12$.

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$\begin{vmatrix} x^2+y^2&x&y&1\\ 5^2+7^2&5&7&1\\ 6^2+6^2&6&6&1\\ 2^2+(-2)^2&2&-2&1 \end{vmatrix}=0$

$-12(x^2+y^2)+48x+72y+144=0$

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