Finding the equation of a circle given the tangent line and a point on that line.

I've solved some of them and the rest I don't really understand how to go about.

1. How do I find the equation of the circle if it is tangent to the line -3x+2y+5 = 0 at the point (1, -1)?

I'm stumped on this one since I don't know how I'd be able get any of the details through equations. I don't have the center point so I can't use the perpendicular distance formula/look for perpendicular line by finding the inverse of the slope (or can I? I'm not sure.)

I'm still quite confused about the circle. And I'm not really sure if the concepts I'm using are correct or not. I'm not sure if I can do this and that for this problem and that problem, but I'm slowly learning. I don't have a proper background on it since I'm just searching willy nilly.

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If you guys don't mind I have follow-ups. You don't need to give me the answers, it would help if you can explain what I'm supposed to do and how to go about doing it (Beginner level formulas for this, I guess).

1. Given circle is tangent to the line -x+y+4 = 0 at point (3, -1) and the circle's center is on the line x + 2y -3 = 0, how will I find the equation of the circle?

2. **Is there a faster way to find out the equation of the circle inscribed in the triangle?**

Take this for example: SOLUTION: Find the equation of the circle that is inscribed in the triangle whose sides lie on the lines 3x+y=5,x-3y=1,and x+3y=-7.

Also in that page why is it that there are only four solutions to the system. Can't the equations on the left side be negative as well?

Here's another example which is shorter but I totally don't understand how she figured out the signs.

SOLUTION: find the equation of the circle inscribed in a triangle whose sides are 3x+y-5=0, x+3y-1=0, x-3y+7=0.

*Thanks for the help guys. It's greatly appreciated.*

Re: Finding the equation of a circle given the tangent line and a point on that line.

Hello, RandomInquirer!

Quote:

2. Is there a faster way to find out the equation of the circle inscribed in the triangle?

Example: Find the equation of the circle that is inscribed in the triangle whose sides

lie on the lines: .$\displaystyle 3x+y\:=\:5,\;\;x-3y\:=\:1,\;\;x+3y\:=\:-7.$

The distance from a point $\displaystyle (x_1,y_1)$ to a line $\displaystyle Ax + By + C \:=\:0$

. . is given by: .$\displaystyle d \;=\;\frac{|Ax_1 + By_1+C|}{\sqrt{a^2+B^2}} $

We want a point $\displaystyle (h,k)$ which is equidistant from: $\displaystyle \begin{Bmatrix}3x+y-5 &=& 0 \\ x-3y - 1 &=& 0 \\ x+3y+7 &=& 0 \end{Bmatrix}$

$\displaystyle \text{We have: }\;d \;=\;\underbrace{\frac{3h+k-5}{\sqrt{10}}}_{[1]} \;=\;\underbrace{\frac{h-3k-1}{\sqrt{10}}}_{[2]} \;=\;\underbrace{\frac{h+3k+7}{\sqrt{10}}}_{[3]}$

$\displaystyle [2] = [3]\!:\;h-3k-1 \:=\:h+3k+7 \quad\Rightarrow\quad -6k \:=\:8 \quad\Rightarrow\quad k \,=\,\text{-}\tfrac{4}{3}\;\;[4]$

$\displaystyle [1] = [2]\!:\;\;3h+k-5 \:=\: h-3k-1 \quad\Rightarrow\quad h \:=\:2 - 2k\;\;[5]$

Substitute [4] into [5]: .$\displaystyle h \:=\:2-2\left(\text{-}\tfrac{4}{3}\right) \quad\Rightarrow\quad h \:=\:\tfrac{14}{3}$

The center of the circle is: .$\displaystyle \left(\frac{14}{3},\;\text{-}\frac{4}{3}\right)$

The radius is: .$\displaystyle [1]\;\;r \;=\;\frac{3(\tfrac{14}{3}) + (\text{-}\frac{4}{3}) - 5}{\sqrt{10}} \;=\;\frac{23}{3\sqrt{10}}$

The equation of the circle is: .$\displaystyle \left(x-\frac{14}{3}\right)^2 + \left(y + \frac{4}{3}\right)^2 \;=\;\frac{529}{90} $

1 Attachment(s)

Re: Finding the equation of a circle given the tangent line and a point on that line.

I used an online circle graph application to compare your answer to the answer given on the hyperlinked page and your answer ended up being the red circle right next to the inscribed circle.

Attachment 30420

From what I gathered, the circle with the smallest radius/diameter is the inscribed circle. Is this true for all or is there some sort of special case where it's not? I can't think of a case where it is false but that doesn't mean that somebody else has not.

Re: Finding the equation of a circle given the tangent line and a point on that line.

Hello again, RandomInquirer!

Quote:

1. Given circle is tangent to the line $\displaystyle -x+y+4 \,=\, 0$ at point $\displaystyle P(3, -1)$

and the circle's center is on the line $\displaystyle x + 2y -3 \,=\, 0,$

find the equation of the circle.

A tangent to the circle is perpendicular to the radius at the point of tangency.

The tangent is: $\displaystyle y \,=\,x-4$; its slope is $\displaystyle 1$.

The slope of the radius is $\displaystyle \text{-}1.$

Hence, the center lies on the line: .$\displaystyle y+1 \,=\,\text{-}1(x-3) \quad\Rightarrow\quad x+y \,=\,2$

The center also lies on the line: .$\displaystyle x + 2y \,=\,3$

Solve the system.

. . The center is $\displaystyle C(1,1).$

The radius is: .$\displaystyle CP \:=\:\sqrt{(3-1)^2 + (-1-1)^2} \:=\:\sqrt{8}$

Therefore: .$\displaystyle (x-1)^2 + (y-1)^2 \:=\:8$

Re: Finding the equation of a circle given the tangent line and a point on that line.

Thank you for your help. It's much appreciated!

2 Attachment(s)

Re: Finding the equation of a circle given the tangent line and a point on that line.

Hi,

Rarely does Soroban make an error, but he did this time; nobody bats a thousand. He didn't choose the right sign for the distance of (h,k) to 3x+y-5=0. Your first link shows a correct solution, but the second link also has an error. I don't see any easy way to solve this problem The two attachments try to explain:

Attachment 30439

Attachment 30440

Re: Finding the equation of a circle given the tangent line and a point on that line.

Quote:

Originally Posted by

**RandomInquirer** I've solved some of them and the rest I don't really understand how to go about.

1. How do I find the equation of the circle if it is tangent to the line -3x+2y+5 = 0 at the point (1, -1)?

All circles have equations of the form $(x-h)^2 + (y-k)^2 = r^2$. Using implicit differentiation, we can find the derivative: $\dfrac{dy}{dx} = \dfrac{h-x}{y-k}$

The line $-3x+2y+5=0$ has slope $\dfrac{3}{2}$

Plugging that in, you get $\dfrac{3}{2} = \dfrac{h-1}{-1-k}$

Additionally, you know $(1,-1)$ is a point on the circle, so $(1-h)^2 + (-1-k)^2 = r^2$.

Solving for $k$ in the first equation, we get: $k = -\dfrac{2h+1}{3}$

Plugging into the second equation, we get:

$(1-h)^2 + \left(\dfrac{2h-2}{3}\right)^2 = r^2$

$r^2 = \dfrac{13}{9}(1-h)^2$

Thus, the circle's formula is

$(x-h)^2 + \left(y+\dfrac{2h+1}{3}\right)^2 = \dfrac{13}{9}(1-h)^2$ where $h \in \mathbb{R}$ (it gives you an infinite family of circles).