algebra-precalculusanalytic geometrygeometrylinear algebramatrices
How to find the equation of a circle which passes through these points $(5,10), (-5,0),(9,-6)$
using the formula
$(x-q)^2 + (y-p)^2 = r^2$.
I know i need to use that formula but have no idea how to start, I have tried to start but don't think my answer is right.
Recall the:
Distance Formula:
The distance between the points $(x_1,y_1)$ and $(x_2,y_2)$ is $$ D=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} $$
Example:
The radius of your circle is the distance between the points $(-1,4)$ and $(3,-2)$. Using the Distance Formula: $$ D= \sqrt{ \bigl(3-(-1)\bigr)^2 + (-2-4)^2 } = \sqrt{ 4^2 + (-6)^2 } = \sqrt{ 16+36 } = \sqrt{ 52 } . $$
What is the equation of the circle?
It is important to realize the "equation of the circle" is: a point $(x,y)$ is on the circle if and only if the coordinates of the point $x$ and $y$ satisfy the equation. So, how to get the equation? What is the relationship between the $x$ and $y$ coordinates of a point on the circle?
Well, let $(x,y)$ be a point on the circle. The big idea is:
$$ \text{The distance from the point }(x,y)\text{ to the center }(-1,4)\text{ is }\sqrt{52}.$$
So, using the distance formula (with $(x_2,y_2)=(x,y)$ and $(x_1,y_1)=(-1,4)$) , it follows that
$$
\sqrt{52}=\sqrt{\bigl(x-(-1)\bigr)^2 +(y-4)^2}.
$$
Or $$ 52=(x+1)^2 +(y-4)^2. $$
The shortcut would be to just use the following formula (But it's important to realize why you'd use it and where it comes from):
Equation of a Circle
The equation of the circle with center located at $(a,b)$ and with radius $r$ is $$ r^2=(x-a)^2 +(y-b)^2 $$ Note that the radius squared is on the left-hand side of the equation.
$$\left | \begin{matrix} x^2 + y^2 + z^2 & x & y & z & 1 \\ x_1^2 + y_1^2 + z_1^2 & x_1 & y_1 & z_1 & 1 \\ x^2_2 + y_2^2 + z_2^2 & x_2 & y_2 & z_2 & 1 \\ x_3^2 + y_3^2 + z_3^2 & x_3 & y_3 & z_3 & 1 \\ x_4^2 + y_4^2 + z_4^2 & x_4 & y_4 & z_4 & 1 \end{matrix} \right | = 0 $$
i.e. $-128(x^2+y^2+z^2-4x-\frac{29}{2}y-4z)=0$
Best Answer
Suppose the points are $A,B,C$. Then intersect the equations of perpendicular bisectors of $AB$ and $BC$. This is the center of the desired circle. (with your notation $(p,q)$)
Now calculate the distance between $(p,q)$ and $A$. Now $r$ is also found.