# Equation of the circle?

1. Jul 9, 2007

I can't understand the different conditions given and from that equation of the circle is found.For example.
How equation of the circle can be found ? if two points on circle are given and equation of line tangent to the circle is given?
thank you.

2. Jul 9, 2007

### CompuChip

Well, you know the general equation for a circle don't you? If not, try to derive it from what you know (namely, they consist of points which have a fixed distance to -- say -- the origin, and you know the distance of a point to the origin by Pythagoras law, then you can shift the whole thing to the actual center of the circle).
This equation contains two coordinates (e.g. x and y) and three unknowns (namely the coordinates of the centre point and the radius). So if you know two points on the circle, you could plug them in and find two of them.
To get the tangent line, you can rewrite the formula to y = f(x) with f a function that depends on x and the radius, and differentiate it.

That's all I'm going to say now, I think you should think about this and come up with some trials, because I don't know exactly where your problem is.

P.S. Make pictures! Try drawing the points and the given tangent line, or just draw a circle and see what you can derive by looking at the picture.

3. Jul 9, 2007

### mathwonk

you could try to use the two points to get an idea of where the center is, since it is equidistant from both. then recall the relation between the center of a circle and the tangent line at a point of the circle. this gives a geometric approach.

4. Jul 9, 2007

### threetheoreom

Yeah i was thinking of telling him that but what if the points that are giving are not equidistant what if for instance they are next to to each other, or should we assume that must precal questions will give equidistant points.

Last edited: Jul 9, 2007
5. Jul 9, 2007

### threetheoreom

yeah give an example of a problem you need to solve.

6. Jul 9, 2007

### HallsofIvy

I think you are misunderstanding mathwonk's point. It doesn't make sense to say that two points are "equidistant". You may be thinking of the case where the two points are ends of a single diameter. Mathwonk said that the center of the circle is equidistant from both the given points on the circle. In particular, the center is on the pependicular bisector of a line segment (chord) between any two points on the circle.

7. Jul 10, 2007

Ok The example is that I am taking fro my text book.
Find an equation of the circle passing through the points A(1,2) and B(1,-2) and touching to the line x+2y+5=0

8. Jul 10, 2007

### Werg22

Well the center obviously lies along the y axis, lucky you. For your convenience write x+2y+5=0 as y = (-x - 5) / 2. The slope of the line perpendicular to this one is 2, so we have y = 2x + b. Now the distance between (0, b) and the intersection of 2x + b with (-x- 5)/ 2 needs to be expressed as a function of b. This done, express the distance between one of the given points and (0, b) as a function of b, equalize the two functions and solve for b.

(Editted by HallsofIvy to change y= -x- 5/2to (-x-5)/2)

Last edited by a moderator: Jul 11, 2007