algebra-precalculus
My problem is asking me to find the equation of a circle that "passes through (-2,1). Does this mean that (-2,1) is the center? Or it is just a point on the circle?
The entire problem:
Find the standard equation of the circle passing through (-2,1) and tangent to the line 3x-2y=6 at the point (4,3). Sketch. (Hint: The line through the center of the circle and the point of tangency is perpendicular to the tangent line.)
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.
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
"Passing through" means that it is on the circle i.e. it is one of the points on the circle.
How to solve:
You are given a point on the circle, $(-2, 1)$, a tangent line $3x - 2y = 6$ and a tangent point $(4, 3)$.
Find the slope of the tangent line:
$3x - 2y = 6$
$-2y = -3x + 6$
$y = \frac{3}{2}x + 3$
The slope is $\frac{3}{2}$.
Recall that the slope of a perpendicular line is the negative reciprocal of the original line. In other words, if $m_1$ is the slope of a line, and $m_2$ is the slope of the perpendicular line, then $m_1 * m_2 = -1$.
The radius must be perpendicular to the tangent line. Using the above formula, let $m_1$ be equal to the slope of the tangent line, $\frac{3}{2}$. Then $m_2$, the slope of the radius, can be found like so:
$\frac{3}{2} * m_2 = -1$
$m_2 = -\cfrac{1}{\left(\cfrac{3}{2}\right)}$
$m_2 = -1 * \frac{2}{3}$
$m_2 = -\frac{2}{3}$
Using point-slope form, derive an equation of a line that the centre is on.
$y - y_0 = m(x - x_0)$
$y - 3 = -\frac{2}{3}(x - 4)$
$y = -\frac{2}{3}x - \frac{8}{3} + 3$
$y = -\frac{2}{3} + \frac{1}{3}$
Let the coordinates of the center be $(a, b)$. Then the equation of the circle is $(x-a)^2 + (y-b)^2 = r^2$
Remember that $b = -\frac{2}{3}a + \frac{1}{3}$
Substitute $b$ for the expression above, and plug in the point $(-2, 1)$ into the equation.
$(-2-a)^2 + (1-(-\frac{2}{3}a + \frac{1}{3}))^2 = r^2$
$a^2 + 4a + 4 + (\frac{2}{3}a - \frac{2}{3})^2 = r^2$
$a^2 + 4a + 4 + \frac{4}{9}a^2 - \frac{8}{9}a + \frac{4}{9} = r^2$
$\frac{13}{9}a^2 + \frac{28}{9}a + \frac{40}{9} = r^2$
Do the same thing, but instead, plug in the point $(4, 3)$ into the equation. To make this faster, $(4-a)^2 + (3-(\frac{2}{3}a + \frac{1}{3}))^2$ evaluates to $a^2 - 8a + 16 + (-\frac{2}{3}a + \frac{8}{3})^2$, which equals $a^2 - 8a + 16 + \frac{4}{9}a^2 - \frac{32}{9}a + \frac{64}{9}$. Further simplification yields $\frac{13}{9}a^2 - \frac{40}{9}a + \frac{80}{9}$.
Now you have two equations in two variables. This creates a system.
$\left\{ \begin{array}{rcl} \frac{13}{9}a^2 + \frac{28}{9}a + \frac{40}{9} = r^2 \\ \frac{13}{9}a^2 - \frac{40}{9}a + \frac{80}{9} = r^2 \\ \end{array} \right.$
Multiply both equations by $9$ to get rid of the fractions. This allows for easier work.
$\left\{ \begin{array}{align} 13a^2 + 28a + 40 = 9r^2 \\ 13a^2 - 9a + 80 = 9r^2 \\ \end{array} \right.$
Equate the expressions and solve.
$13a^2 + 28a + 40 = 13a^2 - 9a + 80$
$37a - 40 = 0$
$37a = 40$
$a = \frac{40}{37}$
Plug in the value of $a$ into the equation $b = -\frac{2}{3}a + \frac{1}{3}$ defined above.
$b = -\frac{2}{3} * \frac{40}{37} + \frac{1}{3}$
$b = -\frac{80}{111} + \frac{1}{3}$
$b = -\frac{80}{111} + \frac{37}{111}$
$b = -\frac{57}{111}$
The center of the circle is at $(\frac{40}{37}, -\frac{57}{111})$.
To find $r^2$, use the distance formula, but square both sides so that the square root is removed. Allow me to demonstrate:
Original distance formula: $D = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$, where $D$ is the radius $r$, and $(x_1, y_1)$ and $(x_2, y_2)$ are any two given points.
Squaring both sides gives $(x_1 - x_2)^2 + (y_1 - y_2)^2 = r^2$, which looks a lot like the equation of a circle (it actually is not).
Plug in the points $(\frac{40}{37}, -\frac{57}{111})$ and $(-2, 1)$ into the formula.
$(\frac{40}{37} + 2)^2 + (-\frac{57}{111} - 1)^2 = r^2$
$(\frac{114}{37})^2 + (-\frac{168}{111})^2 = r^2$
$\frac{12996}{1369} + \frac{28224}{12321} = r^2$
$\frac{198762372}{16867449} = r^2$
The equation of the circle is:
$(x - \frac{40}{37})^2 + (y + \frac{57}{111})^2 = \frac{198762372}{16867449}$
I might have made a calculation mistake / error, but you get the idea of how to do it. I hope this answer helped you!
EDIT: Yes I did make a calculation error when I was solving the system, but you get the point. DO NOT use my answer for your assignment / homework, because it is wrong.