What is the equation of the circle

Question

I've been experiencing a difficulties in answering this. I hope someone will help me in solving this

Find the equation of the circle through the points $(2,8),(7,3)$ and $(-2,0)$.

Suan Welcome to MathSE. Can you show your attempts to solve this problem? For example, do you know the equation of the circle? Have you solved a similar problem? If yes, what did you learn? — , Oct 10 '20 at 16:36
Or if you know the general equation of the circle in 2D in terms of its center and $r$, you have $3$ variables and $3$ equations to solve. — Math Lover, Oct 10 '20 at 16:36
I've been doing the elimination method, but still I couldn't get the right answer — Suan Suan, Oct 10 '20 at 16:43
If you were given a scale, compass and a pencil and 3 arbitrary points on a paper, How would you draw the circle? if you can answer this, applying the same algorithm would work as well (Hint: find the intersection of perpendicular bisectors of the lines formed by any 2 randomly chosen points) — Anindya Prithvi, Oct 10 '20 at 16:49
@Suan I obtained the equation $(x-2)^{2}+(y-3)^{2}=5^{2}$ using hint of user. — , Oct 10 '20 at 16:55
@Anindya Prithvi, You're right. I deleted that answer. When I read the problem I accidentally changed the point $(2,8)$ to $(2,7)$ and thus solved the problem. And I corrected that type and got $(x-2)^{2}+(y-3)^{2}=5^{2}$. — , Oct 10 '20 at 16:58
The general equation of a circle is $x^2+y^2+2gx+2fy+c$ =0 Try substituting the points and find g,f and c — Shashaank, Mar 13 '23 at 09:53

score 3 · Answer 1 · edited Oct 10 '20 at 17:42

3

Let $A(2,8),B(7,3)$ and $C(-2,0)$

Using, "$\perp$ line from centre bisects chord"

Midpoint of AC is $M_{AC}=(0,4)$

Midpoint of BC is $M_{BC}=\left(\dfrac{5}{2},\dfrac{3}{2}\right)$

Slope of line the lines AC and BC are $m_{AC}=2$, $m_{BC}=\frac{1}{3}$

and slope of perpendicular bisectors of AC and BC $m_1=-\frac{1}{2}$,$m_2=-3$,

and equation of perpendicular bisectors are $y=-\frac{1}{2}x+4$ and $y=-3x+9$, respectively

Coordinates of intersection of perpendicular bisectors are $O(2;3)$ that is centre of the circle.

Radius is $r=OC=5$ or $r=OA=OB=OC$

$(x-2)^2+(y-3)^2=25$

edited Oct 10 '20 at 17:42

Anindya Prithvi

1,194

answered Oct 10 '20 at 17:08

Lion Heart

7,073

1

(+1)Thanks for including this – Anindya Prithvi Oct 10 '20 at 17:14
Thanks a a lot. – Suan Suan Oct 10 '20 at 17:26
You are welcome – Lion Heart Oct 10 '20 at 17:45
How did you get the 9 in the y=-3x+9? – Suan Suan Oct 10 '20 at 18:00
Using the point $M_{BC}$ and $m_2$ – Lion Heart Oct 10 '20 at 18:03

user · Answer 2 · 2020-10-10T17:13:21.483

2

HINT

Let start from the general equation

$$(x-x_C)^2+(y-y_C)^2=R^2$$

and plug in the values $(x,y)$ for the given points to obtain three equations in the three uknowns $x_C$, $y_C$ and $R$.

We obtain the system

$2x_1x_C+2y_1y_C-x_1^2-y_1^2=x_C^2+y_C^2-R^2$
$2x_2x_C+2y_2y_C-x_2^2-y_2^2=x_C^2+y_C^2-R^2$
$2x_3x_C+2y_3y_C-x_3^2-y_3^2=x_C^2+y_C^2-R^2$

that is

$2x_1x_C+2y_1y_C-x_1^2-y_1^2=2x_2x_C+2y_2y_C-x_2^2-y_2^2$
$2x_2x_C+2y_2y_C-x_2^2-y_2^2=2x_3x_C+2y_3y_C-x_3^2-y_3^2$

from which we can find $x_C$ and $y_C$ and then $R$.

We can check directly the solution $(x-2)^2+(y-3)^2=25$ you have obtaind as follows

$(2,8) \to (2-2)^2+(8-3)^2=25$
$(7,3) \to (2-7)^2+(3-3)^2=25$
$(-2,0) \to (-2-2)^2+(0-3)^2=25$

edited Oct 10 '20 at 17:13

answered Oct 10 '20 at 16:36

user

154,566

I've been doing that lately, but still I couldn't guess the right answer. Could please show the solution? – Suan Suan Oct 10 '20 at 16:38
@SuanSuan You should indeed show your work to better understand your difficulties. – user Oct 10 '20 at 16:42
I used online calculator in solving the equation and I get (x-2)^2+(y-3)^2= 25 – Suan Suan Oct 10 '20 at 16:50
@SuanSuan You can check it plugging in the given values. – user Oct 10 '20 at 16:52
@Suan Suan. Yes, that's the answer for your problem. – Oct 10 '20 at 17:00
I used online calculator in solving the equation and I get (x-2)^2+(y-3)^2= 25 .
But when I'm solving it using this method, this is what I get. (2,8) (7,3) (-2,0)

x^2+y^2+Dx+Ey+F=0 For (2,0) 2D+8E+F=68 For (7,3) 7D+3E+F=58 For (-2,0) -2D+F=4 (b) = 7D+3E+F=58 (a) = -2D-8E-F=-68 + (d)= 5D-5E=-10

(c) -2D+F= 4 (a) -2D-8E-F= -68 + (e) -4D-8B=-64

4 (d) 20D-20E= -40 5 (e) -20D-40E= -320 + -60E= -360 or equal E= 6

I substitute 6 in the (d), 5D+5(6)=-10, equal to D= 4 I substitute 4 in the (c), -2(4)+F=4 , equal to 12
– Suan Suan Oct 10 '20 at 17:25
Let update your question with your full calculations, I’ll take a look to that later! – user Oct 10 '20 at 17:27

score 2 · Accepted Answer · answered Oct 10 '20 at 18:04

The question can be restated as

Given triangle $ABC$ with coordinates of vertices $A=(-2,0)$, $B=(7,3)$, $C=(2,8)$, find the equation of its circumscribed circle.

First, find the squares of the side lengths of $\triangle ABC$: \begin{align} a^2&=50 ,\quad b^2= 80 ,\quad c^2=90 \tag{1}\label{1} . \end{align}

Second, the coordinates of the circumcenter is known to be found as

\begin{align} O&= \frac{a^2(b^2+c^2-a^2)\cdot A+b^2(a^2+c^2-b^2)\cdot B+c^2(b^2+a^2-c^2)\cdot C} {a^2(b^2+c^2-a^2)+b^2(a^2+c^2-b^2)+c^2(b^2+a^2-c^2)} \\ &=\tfrac1{12}\cdot(5\,A+4\,B+3\,C) =\tfrac1{12}\cdot(5\,(-2,0)+4\,(7,3)+3\,(2,8)) =(2,3) \tag{2}\label{2} . \end{align}

Third, find the radius: \begin{align} R&=|O-A|=|O-B|=|O-C|=5 \tag{3}\label{3} . \end{align}

Hence, the equation of the circle is

\begin{align} (x-2)^2+(y-3)^2&=25 \tag{4}\label{4} . \end{align}

score 1 · Answer 4 · answered Oct 10 '20 at 16:58

1

Satisfy the general equation of circle $$x^2+y^2+2gx+2fy+c=0$$ with these three points get linear equations in $f,g,c$ and solve them.

answered Oct 10 '20 at 16:58

Z Ahmed

43,235

What is the equation of the circle

4 Answers4