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### Analytic Geometry

Date: 08/31/97 at 22:49:47
From: Ann
Subject: Analytic geometry

How do I find the standard equations of the circles that pass through
(2,3) and are tangent to both the lines 3x - 4y = -1 and 4x + 3y = 7?

I have looked at the equations of both lines and see that they are
perpendicular to each other. They intersect at (1,1). I have given the
center of the circle an arbitrary name, say (a,b), and used the
formula for the distance between a point and a line using the same
center and both lines, thinking that then I could set them equal to
each other. I needed to use (2,3) somehow, so I put that in the
formula for the distance between two points using the center (a,b) and
(2,3).

But then in order to get rid of the square root sign you must square
each side, which leaves you with r to the second power. Then I'm
stuck. Am I on the right track? Thanks for your help! This is very
frustrating.

Date: 09/01/97 at 11:26:29
From: Doctor Jerry
Subject: Re: Analytic geometry

Hi Ann,

I think it is a good idea to use the formula for the distance from the
center (a,b) of the circle to one or two of the given lines.

An idea that may help simplify your work can be found by noting that
the center (a,b) of the circle you are looking for lies on the angle
bisector of the two lines. The line with slope 3/4 has an angle of
inclination arctan(3/4). So, since the given lines meet at 90 degrees,
the angle of inclination of the bisection will be t=arctan(3/4)+pi/4.
By taking the tangent of both sides, you can show that the slope of
the line through (a,b) is 7.  So, from your work, b-1 = 7(a-1).

Since (2,3) is on the circle,

(2-a)^2+(3-b)^2 = r^2.

Lastly, using the distance formula, as you mentioned, the distance
from (a,b) to the line 3x-4y = -1 is

r = |3a/4-b+1/4|/(5/4).

You can try both signs for the absolute value. One gives negative r
and so we ignore it. The other gives a nice linear equation connecting
a, b, and r.  A solution to the system is then forthcoming. You should
find r = 5 and (a,b) = (2,8) as the only solution.

-Doctor Jerry,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/

Date: 09/12/98 at 08:48:15
From: Doctor Ujjwal
Subject: Re: Analytic geometry

In general, the above problem will have TWO solutions. You can see this
by making a quick sketch:

Draw two circles of unequal size, so that they either intersect (or are
tangential to) each other. Then draw the two common (external) tangents.

In this figure the tangents are like the two given lines:

L1: 4x + 3y = 7
and    L2: 4x - 3y = 1

One of the points of intersection (or the point of tangency) of the
circles is like the given point

P = (2, 3).

The centers:

C1, C2 : (a, b)

of the two circles are the two solutions we are seeking.

Only if the point P lies on L1 or L2 will the two solutions 'coincide'.
But that is not the case here.

-----------------------

You can proceed as Doctor Jerry has so well written. But notice that
L1 and L2 divide the plane in four quadrants and there are two possible
angle bisectors:

Bisector B1 with slope 7 and
Bisector B2 with slope -1/7

Choose bisector B1 because it passes through the quadrant in which point
P lies.

Since B1 has slope 7 and it must pass through the point of intersection
(1, 1) of L1 and L2, its equation is:

y = 7x - 6

Since (a, b) lies on B1,

b = 7a - 6

Set up the equations as Doctor Jerry has shown and solve for 'a'.

You will get a QUADRATIC equation, with solutions:

a = {2, 6/5}

the corresponding values of 'b' will be

b = {8, 12/5}

Thus the centers of the two circles are:

C1(2, 8) and C2(6/5, 12/5)

distances PC1 = 5 and PC2 = 1 are their radii.

From the centers and radii, the standard forms of the two circles can be
written.

- Doctor Ujjwal Rane

Associated Topics:
College Conic Sections/Circles
High School Conic Sections/Circles

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