# I need math help with tangent lines and circles!?!?

The equation of the circle which is centered at (-4,-6) and which has the line -x+3y+7=0 as a tangent is

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The equation of the circle is (x+4)^2 + (y+6)^2 = r^2 where r is to be determined.

r is just the distance from (-4,-6) to the line x+3y+7= 0

This distance is given by the formula:

d = |ax1+by1+c|/sqrt(a^2+b^2), where a = 1, b= 3 and c = 7 and x1 = -4 and y1 = -6.

So d = r = |(1)((-4) + (3)(-6) + 7|/sqrt(1^2 + 3^2)

r = |-11|/sqrt(10)

So r^2 = 121/10 = 12.1

The equation is (x+4)^2 + (y+6)^2 = 12.1

• For this one you do not know the radius.. You have to find it.

let r = radius -- a constant

let (x2,y2) be the point on the line that is tangent to the circle

Now you need to find the equation of a circle:

(x+4)^2 + (y+6)^2 = r^2

You are given the tangent line:

3y = x-7

y = x/3 - 7/3

You know that a tangent line is always perpendicular to a line from the center of the circle to the point of tangent.

Given the tangent line, find a line that is perpendicular to it which contains the point (-4,-6).

Since the tangent line:

y = x/3 - 7/3 , has slope of 1/3 a perpendicular slope is -3

Now:

(y-y1)= m(x-x1)

(y+6) = (-3)(x+4)

y = -3x -18 -- is a line that is perpendicular to the tangent and contains the mid point of the circle.

Where these two lines intersect is the point where the tangent line has a point on the circle!

Solving for x and y:

x = -47/10

y = -39/10

All that is left to do is find the distance between the midpoint and

the tangent point. This is your r value.

The rest I am sure you can figure out.

EDIT:

I Found r^2 = 49/10 so the equation comes out to:

(x+4)^2 + (y+6)^2 = 49/10

r^2 = 49/10

r = 7/sqrt(10) = 7*sqrt(10)/10 -- this is the most correct way of writing it.

remeber for circle:

y = +/- sqrt((49/10)-(x+4)^2) -6

I graphed out all the lines and it made sense. You can check yourself by using this program and typing in:

http://www.walterzorn.com/grapher/grapher_e.htm

Type in the following (or copy paste):

sqrt((49/10)-(x+4)^2) -6;

-sqrt((49/10)-(x+4)^2) -6;

x/3 -7/3;

-3x -18;

Those are all the graphs

• Distance from point (xo,yo) to line Ax + By + C = 0 is given by

| Axo + Byo + C| / SQRT(A^2 + B^2).

|1*-4 + 3*-6 +7| / SQRT(1^2 + 3^2) = 4.74