Analytic GeometryDate: 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. Now, about the solution: ----------------------- 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 |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/