Finding a circle

[Probability]

OK I have some points. A(5, -3), B(-1, 1) and C(0.2)

I have calculated the slope that passes through AB

y2 - y1 = 1 - (-3) = - ⁴/₆ = - ²/₃
x2 - x1 - 1 - 5


Slope AB = - ²/₃


Coordinates of the midpoints AB


(¹/₂ (x₁+x₂), ¹/₂(y₁ + y₂))

(¹/₂(5 + -1), ¹/₂(-3 + 1)

M_AB = {⁴/₂, -²/₂}

M_BC = {-¹/₂, ³/₂}

Finding the equation of perpendicular bisector AB

M_AB = ²/₃

Equation is;

b_{AB :} y - (-²/₂) = ²/₃(x - ⁴/₂) this implies y = ²/₃x - 3

At this point I am unsure if an error is present, the reason being is that in my previous thread Parametric Equations, I concluded that y = (3x - 8) / (2)

so if then (3(4/2) - 8)) / (2) = - ²/₂ = - 1, then using the equation above just found;

y = ²/₃x - 3 implies ²/₃ (⁴/₂) - 3 = - 1.66 or - ⁵/₃

Looking at this I am suspecting that the centre of the circle lies at points (2, -1.66)

How am I doing?

 

[Deleted member 4993]

OK I have some points. A(5, -3), B(-1, 1) and C(0.2)

I have calculated the slope that passes through AB

y2 - y1 = 1 - (-3) = - ⁴/₆ = - ²/₃
x2 - x1 - 1 - 5


Slope AB = - ²/₃


Coordinates of the midpoints AB


(¹/₂ (x₁+x₂), ¹/₂(y₁ + y₂))

(¹/₂(5 + -1), ¹/₂(-3 + 1)

M_AB = {⁴/₂, -²/₂}

M_BC = {-¹/₂, ³/₂}

Finding the equation of perpendicular bisector AB

M_AB = ²/₃

Equation is;

b_{AB :} y - (-²/₂) = ²/₃(x - ⁴/₂) this implies y = ²/₃x - 3

At this point I am unsure if an error is present, the reason being is that in my previous thread Parametric Equations, I concluded that y = (3x - 8) / (2)

so if then (3(4/2) - 8)) / (2) = - ²/₂ = - 1, then using the equation above just found;

y = ²/₃x - 3 implies ²/₃ (⁴/₂) - 3 = - 1.66 or - ⁵/₃

Looking at this I am suspecting that the centre of the circle lies at points (2, -1.66)

How am I doing?

What does M_AB stand for in those two equations?

 

[Probability]

What does M_AB stand for in those two equations?

Midpoint of AB

Midpoint of BC

 

[Deleted member 4993]

Midpoint of AB

Midpoint of BC

But you have defined both as M_AB

Which one is mid-point of BC?

.

 

[Probability]

.

m_bc = {-¹/₂, ³/₂}

 

[soroban]

Hello, Probability!

Hee's another approach . . .

Find the equation of the circle through: .\(\displaystyle A(5,\text{-}3),\;B(\text{-}1,1),\;C(0,2)\)

The center \(\displaystyle O(x,y)\) is equidistant from \(\displaystyle A,B,C.\)

We have: .\(\displaystyle \begin{Bmatrix}\overline{AO}^2 &=& (x-5)^2 + (y+3)^2 & [1] \\ \overline{BO}^2 &=& (x+1)^2 + (y-1)^2 & [2] \\ \overline{CO}^2 &=& x^2 + (y-2)^2 & [3] \end{Bmatrix}\)


\(\displaystyle [1] = [3]\!:\;x^2 - 10x + 25 + y^2 + 6y + 9 \:=\: x^2 + y^2 - 4y + 4 \)

. . . . . . . \(\displaystyle \text{-}10x + 10y \:=\:\text{-}30 \quad\Rightarrow\quad \color{blue}{x - y \:=\:3}\)


\(\displaystyle [2] = [3]\!:\;x^2+2x+1 + y^2 - 2y + 1 \:=\: x^2+y^2-4y+4\)

. . . . . . . \(\displaystyle 2x + 2y \:=\:2 \quad\Rightarrow\quad \color{blue}{x + y \:=\:1}\)


Solve the system of equations: .\(\displaystyle O(2,\text{-}1)\)

Then: .\(\displaystyle r^2 \:=\:\overline{CO}^2 \:=\: 2^2 + (\text{-}1-2)^2 \:=\:13\)


Therefore: .\(\displaystyle (x-2)^2 + (y+1)^2 \:=\:13\)

 

[Probability]

Hello, Probability!

Hee's another approach . . .


The center \(\displaystyle O(x,y)\) is equidistant from \(\displaystyle A,B,C.\)

We have: .\(\displaystyle \begin{Bmatrix}\overline{AO}^2 &=& (x-5)^2 + (y+3)^2 & [1] \\ \overline{BO}^2 &=& (x+1)^2 + (y-1)^2 & [2] \\ \overline{CO}^2 &=& x^2 + (y-2)^2 & [3] \end{Bmatrix}\)


\(\displaystyle [1] = [3]\!:\;x^2 - 10x + 25 + y^2 + 6y + 9 \:=\: x^2 + y^2 - 4y + 4 \)

. . . . . . . \(\displaystyle \text{-}10x + 10y \:=\:\text{-}30 \quad\Rightarrow\quad \color{blue}{x - y \:=\:3}\)


\(\displaystyle [2] = [3]\!:\;x^2+2x+1 + y^2 - 2y + 1 \:=\: x^2+y^2-4y+4\)

. . . . . . . \(\displaystyle 2x + 2y \:=\:2 \quad\Rightarrow\quad \color{blue}{x + y \:=\:1}\)


Solve the system of equations: .\(\displaystyle O(2,\text{-}1)\)

Then: .\(\displaystyle r^2 \:=\:\overline{CO}^2 \:=\: 2^2 + (\text{-}1-2)^2 \:=\:13\)


Therefore: .\(\displaystyle (x-2)^2 + (y+1)^2 \:=\:13\)

Thanks.

It will take some studying for me to work it out.