algebra-precalculuscircles
No idea how to do this, I used to have these conic shapes committed to memory but I forget them already.
I am supposed to find an equation for the circle that has center $(-1, 4)$ and passes through $(3, -2)$.
I tried graphing it, but it didn't help since the point was not in a straight line with the center and I have no idea what the diameter is.
First, find the center $(a, b)$. It is equidistant from the two points, so $$ \sqrt{a^2 + (b - 2)^2} = \sqrt{(a - 6)^2 + (b - 6)^2} $$ or $$ a^2 + (b - 2)^2 = (a - 6)^2 + (b - 6)^2. $$ Now, since the center is on the line $x - y = 1$, we know that $a - b = 1$, or $$ a = b + 1. $$ Substitute this into the quadratic equation above and simplify: $$ \begin{align} (b + 1)^2 + (b - 2)^2 &= ((b + 1) - 6)^2 + (b - 6)^2 \\ (b^2 + 2b + 1) + (b^2 - 4b + 4) &= (b^2 - 10b + 25) + (b^2 - 12b + 36) \\ 2b^2 - 2b + 5 &= 2b^2 - 22b + 61 \\ 20b &= 56 \\ b &= \frac{14}{5}. \end{align} $$ Therefore, $$ a = \frac{14}{5} + 1 = \frac{19}{5}. $$ What is the radius $r$? Well the square of the radius is $$ \begin{align} r^2 &= a^2 + (b - 2)^2 \\ &= \left( \frac{19}{5} \right)^2 + \left( \frac{4}{5} \right)^2 \\ &= \frac{377}{25}. \end{align} $$ Putting this together, we can write the equation of the circle as $$ \left( x - \frac{19}{5} \right)^2 + \left( y - \frac{14}{5} \right)^2 = \frac{377}{25} $$ or, by multiplying by $25$ to clear denominators, $$ (5x - 19)^2 + (5y - 14)^2 = 377. $$ Here's a picture.
You would rewrite this equation into the standard form for the equation of a circle. You would do this by completing the squares. For instance,
$x^2+y^2-4x+10y+26=0 \implies (x^2-4x+4)-4+(y^2+10y+25)-25+26=0 \implies (x-2)^2+(y+5)^2 = 3 \implies$
center of the circle is the point $(2,-5)$. Based on your insufficient information (no x-intercept is provided), there are infinitely many lines that pass through the center of the circle. We would have to know the x-intercept in order to solve this problem fully.
Best Answer
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.
The following may help: