# Find the equation of the circle passing …

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).

• You get a system of 3 equations and 3 unknowns that you can solve. – Markus Oct 1 '16 at 0:29
• @ Markus, I get problem in solving those equations. Could not get the answer. – pi-π Oct 1 '16 at 0:37
• Using LinAlg and a computer to row reduce, I get $g=-7/3,f=-10/3,c=-4$. – Bobson Dugnutt Oct 1 '16 at 0:45
• "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. – fleablood Oct 1 '16 at 1:08

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 $(2- 2)^2+ (-2- 3)^2= 0+ 25= 25$.

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 $g=-7/3$.

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?

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 $c=-12$.

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

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