circle, tangent, point problems?????

How do you solve these problems?

1) find standard equation of circles that pass thru (2,3) and are tangent to both lines 3x-4y=-1 and 4x+3y=7

2 find standard eqations of circles that have centers on 4x+3y=8 and are tangent to both the line x+y=-2 and 7x-y=-6

3) find eqations of lines thru (4,10) and tangent to circle x^2+y^2-4y-36=0

these problems come at the beginning of calculus 1???? help!!!!!!!!

Re: circle, tangent, point problems?????

Quote:

Originally Posted by

**sluggerbroth** How do you solve these problems?

1) find standard equation of circles that pass thru (2,3) and are tangent to both lines 3x-4y=-1 and 4x+3y=7

2 find standard eqations of circles that have centers on 4x+3y=8 and are tangent to both the line x+y=-2 and 7x-y=-6

3) find eqations of lines thru (4,10) and tangent to circle x^2+y^2-4y-36=0

Hint: Tangents have the same gradient as curves at the point at which they are tangent.

Re: circle, tangent, point problems?????

these problems come at the beginning of calculus 1???? help!!!!!!!!

Re: circle, tangent, point problems?????

Quote:

Originally Posted by

**sluggerbroth** these problems come at the beginning of calculus 1???? help!!!!!!!!

For the first question, what is the gradient of each of your tangents?

Re: circle, tangent, point problems?????

One method of solving problem 1 comes from analytic geometry. As you can check, the lines 3x - 4y = -1 and 4x + 3y = 7 are perpendicular and intersect at (1, 1). Let us consider two coordinate systems with the origin (1, 1): one has axes parallel to the original axes (the latter are drawn dashed in the figure below) and the other has the lines 3x - 4y = -1 and 4x + 3y = 7 as axes. The point A(2, 3) in the original (dashed) system has coordinates and in the new (x, y)-system.

https://lh6.googleusercontent.com/-f...0/parabola.png

The circle that touches x'- and y'-axes must have its center O on the angle bisector that has the equation y' = x'. In addition, the distance OC between B and the x'-axis must equal AO. (In this problem, AO happens to be parallel to the x'-axis, but this is a coincidence.) How do we find the coordinates of O? Recall that the locus of points equidistant from the point A and the x'-axis is a parabola. It is easier to find its equation in the system (x'', y') where the y'-axis is shifted so that it goes through A (dashed line). Then . If O has coordinates (x'', y'), then and . So, is the equation of the parabola. The equation of the bisector in the (x'', y')-system is y' = x'' + 2. Solving these two equations allows finding x'' (and hence x') and y' knowing and .

We know the coordinates of A in the (x, y)-system. How do we find the coordinates of the *same point* in the (x', y')-system? Suppose that and are unit vectors along the x'-, y'-axes expressed in terms of x and y. Then

i.e.,

(*)

This allows finding from , which is also needed below. To find from there are several ways. First, one can solve the system (*) for . Second, one can use the formula for the distance of a point from a line . This distance is .

So, here are the instructions.

Step 1. Find the unit vectors and along the x'- and y'-axes.

Step 2. Find the coordinates of A in the (x', y')-system.

Step 3. Find the equation of the parabola in the (x'', y')-system.

Step 4. Find the coordinates of the circle center O in the (x'', y')-system.

Step 5. Convert the coordinates of O into the (x', y')-system and then (x, y)-system using (*).

Step 6. Find the radius OA = OC of the circle and write the circle equation in the (x, y)-system.