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My procedure: First, I solved for the slope of line CA. Afterwards, I solved for its Midpoint. Then I formed its equation. I also applied the same process to line CB. According to my research, I should subtract the first equation to the second equation. I'm confused because the value of my x was a fraction after subtracting. I decided to make it zero. Then, I used it and solved for the value of y. I got 25/4. I solved for the midpoint by the points C and (0,25/4) I got the center (2,13/5) I guess converse of angle in a semicircle is involved with this.

I'm probably wrong with my calculation, so I'm open for any corrections. Please explain your answer in detailed. Thank you.

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  • $\begingroup$ By center, do you mean the circumcentre? $\endgroup$
    – peterwhy
    Jun 1, 2017 at 23:38
  • $\begingroup$ Have you tried using the equation of a circle? Call the center $(h, k)$. What equations can you write? $\endgroup$
    – N. F. Taussig
    Jun 1, 2017 at 23:43
  • $\begingroup$ @peterwhy yes you're right $\endgroup$
    – Janine
    Jun 1, 2017 at 23:45
  • $\begingroup$ @N.F Taussig Point B will be (x-1)+(y-4)=r^2 Point C will be (x+1)+(y-2)=r^2 Lastly, point A will be (x-4)+(y+3)=r^2 $\endgroup$
    – Janine
    Jun 1, 2017 at 23:50
  • $\begingroup$ What do you mean, you had a fraction, so you made it zero? $\endgroup$
    – Doug M
    Jun 1, 2017 at 23:54

5 Answers 5

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If you are comfortable solving three linear equations in three unknowns, let the equation of the circle be $$(x-h)^2+(y-k)^2 = r^2.$$ Substitute points $A$, $B$ and $C$, you will obtain three linear equations with unknowns $h$, $k$ and $(r^2-h^2-k^2)$.

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  • $\begingroup$ I'm sorry our teacher didn't teach us how to solve three linear equations in three unknowns $\endgroup$
    – Janine
    Jun 2, 2017 at 1:28
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$\left(\frac{1-1}{2},\frac{4+2}{2}\right)$ or $(0,3)$ is a middle point of $BC$. $$m_{BC}=\frac{4-2}{1+1}=1.$$ Thus, the center placed in the line: $y-0=-1(x-3)$, which says that $O(x,3-x)$ is our center.

Now, since $OA=OC$, we obtain: $$(x-4)^2+(6-x)^2=(x+1)^2+(1-x)^2$$ or $$x^2-8x+16+36-12x+x^2=x^2+2x+1+x^2-2x+1$$ or $$-20x+52=2,$$ which gives $x=2.5$ and $O(2.5,0.5)$.

From here $R=\sqrt{14.5}$.

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  • $\begingroup$ I understand how you come up with the answer 3-x, but how did you get the x? Sorry just confused $\endgroup$
    – Janine
    Jun 2, 2017 at 1:03
  • $\begingroup$ @Janine I fixed my post. See now, please. $\endgroup$
    – Michael Rozenberg
    Jun 2, 2017 at 1:06
  • $\begingroup$ I still don't understand how you got the x of the center O. Also, why isn't it not (x-4)^2+(y-6)^2=(x+1)^2+(y-1)^2 $\endgroup$
    – Janine
    Jun 2, 2017 at 1:25
  • $\begingroup$ @Janine $y=3-x$ and $OA=OC$. $OA=\sqrt{(x-4)^2+(y+3)^2}=\sqrt{(x-4)^2+(6-x)^2}$... $\endgroup$
    – Michael Rozenberg
    Jun 2, 2017 at 1:45
  • $\begingroup$ I have a question, is it also possible that OB=OA? $\endgroup$
    – Janine
    Jun 4, 2017 at 3:45
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The equation of a circle with center $(h, k)$ and radius $r$ is $$(x - h)^2 + (y - k)^2 = r^2$$ We wish to find an equation of a circle that passes through the points $A(4, -3)$, $B(1, 4)$, and $C(-1, 2)$. Substituting these points into the equation above yields \begin{align*} (4 - h)^2 + (-3 - k)^2 & = r^2 \tag{1}\\ (1 - h)^2 + (4 - k)^2 & = r^2 \tag{2}\\ (-1 - h)^2 + (2 - k)^2 & = r^2 \tag{3} \end{align*} Equating the expressions for $r^2$ in equations 1 and 2 yields $$(4 - h)^2 + (-3 - k)^2 = (1 - h)^2 + (4 - k)^2$$ Expanding and simplifying yields \begin{align*} 16 - 8h + h^2 + 9 + 6k + k^2 & = 1 - 2h + h^2 + 16 - 8k + k^2\\ -6h + 14k & = -8\\ 3h - 7k & = 4 \end{align*} Equating the expressions for $r^2$ in equations 1 and 3 yields $$(4 - h)^2 + (-3 - k)^2 = (-1 - h)^2 + (2 - k)^2$$ Expanding and simplifying yields \begin{align*} 16 - 8h + h^2 + 9 + 6k + k^2 & = 1 + 2h + h^2 + 4 - 4k + k^2\\ -10h + 10k & = -20\\ h - k & = 2 \end{align*} You can solve for the coordinates of the center by solving the system of linear equations \begin{alignat*}{3} 3h & - & 7k & = 4\\ h & - & k & = 2 \end{alignat*} Once you have found the center, you can substitute one of the points for $x$ and $y$ in the equation $(x - h)^2 + (y - k)^2 = r^2$ to determine the radius.

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Find the lengths of the three sides: $BC=\sqrt{8},AB=\sqrt{58},AC=\sqrt{50}$.

Note it is a right angled triangle: $AB^2=AC^2+BC^2$.

In a right angle triangle the center of circumcircle is the midpoint of the hypotenuse. Thus: $$R=\frac {AB}{2}=\frac{\sqrt{58}}{2}.$$

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Here are two different approaches:

This one seems more direct to me.

There exists a point $X = (x,y)$ such that $d(X,A) = d(X,B) = d(X,C)$

$(x-4)^2 + (y+3)^2 = (x-1)^2 + (y-4)^2 = (x+1)^2 + (y-2)^2\\ x^2 - 8x + 16 + y^2 + 6x + 9 = x^2 - 2x + 1 + y^2 - 8y + 16 = x^2 + 2x + 1 + y^2 - 4y + 4$

We can subtract $x^2 + y^2$ across the board.

$ - 8x + 6x + 25 = - 2x - 8y + 17 = 2x - 4y +5$

and consolidate a little bit further

$ - 10x + 10x + 20 = - 4x - 4y + 12 = 0$

and break this up into two equations.

$ - 10x + 10x + 20 = 0\\ - 4x - 4y + 12 = 0$

And solve for $x$ and $y.$

To find the radius.

$r^2 = (x-4)^2 + (y+3)^2 = (x-1)^2 + (y-4)^2 = (x+1)^2 + (y-2)^2$

plug the value of $x,y$ into any of the above.

The approach you suggest works.

Find the lines through the midpoint of $AC$ and $BC,$ with slopes perpendicular to $AC$ and $BC$

Slope of $AC = -1$

Slope of the perpendicular $= 1$

mid-point $= (1.5, -0.5)$

$y+0.5 = x - 1.5\\ -x + y = -2$

Slope of $BC = 1$

Slope of the perpendicular $=-1$

mid-point $= (0, 3)$

$y-3 = - x\\ x + y = 3$

and again, a system of 2 linear equations.

And you will need to apply a distance formula to find the radius.

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