# Find center, radius and a tangent to $x^2+y^2+6x-4y+3=0$

For the circle $x^2+y^2+6x-4y+3=0$ find
a) The center and radius
b) The equation of the tangent line at the point $(-2,5)$

Now, I solved a) and got the equation $$(x+3)^2+(y-2)^2=10$$ with center $=(-3,2)$ and radius $=\sqrt{10}$
Now, I've never learned about the tangent of a circle, but I think that it's a line that touches the outer end of a circle. But I'm not 100% on that. So if anyone can help me out with this, that would be very beneficial. And please do not solve this for me.

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In $(x-3)^2+(y-2)^2=\sqrt10$,shouldn´t that be: $(x-3)^2+(y-2)^2=10$? Because is in this form:$(x-a)^2+(y-b)^2=R^2$.If you say that the radius is $\sqrt 10$,then in the formula you have only $10$. – HipsterMathematician Jul 20 '12 at 20:49
Yes, it should be $${\left( {x + 3} \right)^2} + {\left( {y - 2} \right)^2} = 10$$ – Pedro Tamaroff Jul 20 '12 at 20:53
@AndréNicolas For sure!(forgot the minus signal...) – HipsterMathematician Jul 20 '12 at 20:54

The line tangent to the circle at $(-2,5)$ is the straight line through $(-2,5)$ that is perpendicular to the radius of the circle that runs from the centre of the circle at $(-3,2)$ to the point $(-2,5)$. Find the slope of that radius, use that to get the slope of the tangent line, and you’re on your way.

Edit: I just corrected the $x$-coordinate of the centre. Yours has the wrong sign: $x+3=x-(-3)$, so the correct $x$-coordinate is $-3$.

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So. Find the slope from the radius to $(-2,5)$ and then find the perpendicular slope? – Austin Broussard Jul 20 '12 at 20:55
@Austin: Yes, and then use the point-slope form to find the equation of the tangent line. – Brian M. Scott Jul 20 '12 at 20:56
The slope from the center point to $(-2,5)$ is $\dfrac{3}{1}$. Which would mean the perpendicular slope is $-\dfrac{1}{3}$. So, would that make $y=-\frac{1}{3}x+\frac{13}{3}$? – Austin Broussard Jul 20 '12 at 21:00
@Austin: The straight line with slope $m$ passing through the point $(a,b)$ has the equation $y-b=m(x-a)$; this is called the point-slope form of the equation of the straight line. Once you have this, you can easily transform it to the slope-intercept form (your $y=mx+b$). – Brian M. Scott Jul 20 '12 at 21:03
@AustinBroussard Now you have the slope and a point on the line. As Brian stated earlier, use the point-slope form of the equation of a line to finish the problem. – Code-Guru Jul 20 '12 at 21:05

Let us complete the square. We get $x^2+6*x+9+y^2-4*y+4-9-4+3=0$ or $(x+3)^2+(y-2)^2=10$. Your circle is with center $O(-3,2)$ and radius $r=\sqrt{10}$

For the second question, let us calculate slope, which would be $\text{rise}/\text{run}$ or in our case center is $(-3,2)$ so slope=$(5-2)/(-2+3)=3$ slope of tangent line would be $-1/3$ put this conditions into form of tangent line $y=mx+b$ we get $5=-1/3\cdot(-2)+b$ or $b=5-2/3=13/3$

We get equation of tangent line $y=-1/3\cdot x+13/3$

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For the first part, where did the square root come from? The sign of $3$ is wrong.

You are right that the tangent line is one that touches the circle in one point. The slope of the tangent to a curve is given by the derivative at that point. So rearrange your equation to $y=$ some function of $x$, take $\frac {dy}{dx}$, and evaluate it at $(-2,5)$. When you rearrange the equation you will have a square root and need to think about the sign.

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I never took pre-calculus. I'm skipping from trigonometry to calculus. I know nothing about derivatives. Sorry. – Austin Broussard Jul 20 '12 at 20:54
@AustinBroussard try to take a look at this one – HipsterMathematician Jul 20 '12 at 21:01
The question is clearly tagged as "algebra-precalculus", so a non-calculus solution should be suggested. – Code-Guru Jul 20 '12 at 21:06

Since the question has already been settled nicely, let me demonstrate an unconventional way to find the tangent of a circle through a given point on the circle.

There are three things that allow us to do what we will be doing in the sequel:

1. Any point can be thought of as a "point-circle"; that is, a circle of radius $\textbf 0$.

2. One can always produce the equation for the radical line of two circles.

3. The radical line of two tangent circles is their common tangent line.

Here, now, is the procedure. We are given the point $(-2,5)$ on the circle $x^2+y^2+6x-4y+3=0$. Constructing the equation of a point-circle is a snap (note that as always, we negate the coordinates of the center when inserting them into the equation of a circle):

$$(x+2)^2+(y-5)^2=0$$

which in expanded form, looks like

$$x^2+y^2+4x-10y+29=0$$

From the equivalence of the tangent line and the radical line in this case that I mentioned earlier, all we have to do to find the tangent line to your circle is to find the radical line of the given circle and the point-circle. (Certainly, a point circle lying on another circle is necessarily tangent to it.) The equation of the radical line is easily produced by just subtracting the two given equations; thus,

$$(x^2+y^2+6x-4y+3)-(x^2+y^2+4x-10y+29)=0$$

or, after simplification and solving for $y$,

$$y=\frac{13-x}{3}$$

is the sought radical line, which is also the tangent line of the circle $x^2+y^2+6x-4y+3=0$ through $(-2,5)$.

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