The points $r =(-4, 3),\ s= (7, 4)$ and $t= (8, -7)$ lie on a circle.

Can someone please help me find the centre of the circle?

Okay guys, this is what I've done so far

So you need to find the perpendicular bisector of two chords

So for chord RS y - (1/11)x = 37/11

And for chord ST y + (11)x = -41

Solving through simultaneous equations, I got x =-4, and y =3 as the centre

But the answer has to be x = 2, and y = -2

Can someone explain to me where I went wrong? :/

• Whenever you post any question on this site, do include your own work on the problem and the efforts you have made so that we can provide some "specific" help ..... – user399078 Feb 13 '17 at 14:05
• Please try to make your title more informative.... – user399078 Feb 13 '17 at 14:08
• I've done that, but it still doesn't seem to work – Raj2000 Feb 13 '17 at 14:20
• but you need perpendicular bisector of it – Raj2000 Feb 13 '17 at 14:27
• What is the bisector of $ST$.? – Nosrati Feb 13 '17 at 14:29

$(2,-2)$ of course!

Let $A(-4,3)$, $B(7,4)$ and $C(8,-7)$

Hence, $\measuredangle ABC=90^{\circ}$

• Can you explain this more? – Raj2000 Feb 13 '17 at 14:45
• 2,-2 is just the midpoint of AC, not the centre of the circle – Raj2000 Feb 13 '17 at 14:45
• @Raj2000 Since $\measuredangle ABC=90^{\circ}$, it's the centre of the circle exactly! – Michael Rozenberg Feb 13 '17 at 14:54

Roughly plot those points on paper and notice that the slope of line joining the points $r$ and $s$ is let $m_1$=$\frac{1}{11}$ and that of line joining $s$ and $t$ is let $m_2$=$-11$. This tells us that they are perpendicular lines and meet on the circle at point $s$ at right angle and hence $rt$ is the diametre of circle. So centre of circle is midpoint of $r$ and $t$ =$(2,-2)$.

• Can you spot my mistake? – Raj2000 Feb 13 '17 at 14:22
• Are my simultaneous equations correct? – Raj2000 Feb 13 '17 at 14:22
• I got y +3/2 = 1/11 (x-[15/2]) – Raj2000 Feb 13 '17 at 14:41
• This gives y - 1/11x - 15/22 -3/2 – Raj2000 Feb 13 '17 at 14:41
• This gives y - 1/11x = -24/11 – Raj2000 Feb 13 '17 at 14:41

Find the attache solution as an image file. Solve those system of equation. • Can you help me solve the equations please? – Raj2000 Feb 13 '17 at 14:36
• How do you solve the equations? – Raj2000 Feb 13 '17 at 14:44
• Its $3 \times 3$ equation that you can solve by elimination method. – Sachchidanand Prasad Feb 13 '17 at 22:37

The bisector of $ST$ is $y=\dfrac{1}{11}x-\dfrac{24}{11}$.

The bisector of $RS$ is $y=-11x+20$.

The contact of $ST$ and $RS$ is $(2,-2)$.

Note that the solution of Michael Rozenberg is genius. Plot points and check that $AC^2=AB^2+BC^2$ so $B=90^\circ$ and $AC$ will be the diameter of the circle, then the center of circle is the midpoint of $AC$.

• I just saw that $\frac{1}{11}\cdot(-11)=-1$. – Michael Rozenberg Feb 13 '17 at 14:58

You made two separate but related mistakes. The equation that you have for the perpendicular bisector of $\overline{RS}$ is actually the equation of the line that passes through those two points. You second equation is that of a line parallel to $\overline{ST}$, but that passes through $R$. When you computed their intersection, you of course got back the coordinates of the point $R$.

Without seeing the details of your work, it’s hard to say exactly where you went wrong, but at the very least, it looks like you confused the direction vector of a line with its normal (perpendicular) vector. That is, instead of using the slope $m$ for each of the lines, you ended up getting $-1/m$ instead, which is at right angles to the correct slope.