# Thread: circles given a point and two tangent lines

1. ## circles given a point and two tangent lines

Hi, I have the following question: 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.

My beginning of the solution: let the center of a circle be given by C(c,d).

Then I have the distance from (2,3) to C, given by

$\displaystyle \sqrt{(c-2)^2+(d-3)^2}$

This must be equal to the shortest distance to both the given lines, which is

$\displaystyle d = \frac{|3c - 4d + 1|}{5 }$
and also:
$\displaystyle d = \frac{|4c + 3d - 7|}{5 }$

Here I'm struck. I think this equation is not very helpful:

|3c - 4d + 1| = |4c + 3d - 7|

Can somebody give me a hint how to proceed? (I don't want the solution, I have it in my textbook)

2. If two lines are parallel, the center of any circle having those two lines as tangents lies on the line equidistant from the two lines. if two lines intersect, the center of any circle having those two lines as tangents lies on one of the two bisectors of the angles the lines make when they intersect.

The slopes of these two lines are $\displaystyle \frac{3}{4}$ and $\displaystyle -\frac{4}{3}$. Because those are not the same the lines are not parallel- but they have the nice property that their product is -1 so the two lines are perpendicular. It also will help to know where the two lines intersect.

3. Thanks for your answer, HallsofIvy. I see that the two tangent lines intersect at (1,1). Applying Pythagoras to this point and to the radius gives:
$\displaystyle 4((2-c)^2 + (3-d)^2) = (c-1)^2 + (d-1)^2$
where the right part of the equation is the distance between (1,1) and (c,d)

But this doesn't help me further. I don't know how to obtain the equation of the bisector of the two tangent lines.

4. |3c - 4d + 1| = |4c + 3d - 7|

The above equation gives you
c + 7d = 8 or c = 8 - 7d.
Now

$\displaystyle r^2 = (c - 2)^2 + (d-3)^2 = \frac{(3c - 4d +1)^2}{25}$

Substitute the value c from above in the expression of r^2 and solve the quadratic for d. You will get two values for d. From that find the values of c and proceed.

5. Thanks Sa-ri-ga-ma, but, although you seem more experienced than me, I think that

|3c - 4d + 1| = |4c + 3d - 7| => c = 8 - 7d is not correct. At least, this last equation doesn't fit with the values for c and d I have in the answer. Please correct me if I'm wrong.

In the meantime a friend helped me with the solution:

First, |3c - 4d + 1| = |-(3c - 4d + 1)| = |4d - 3c - 1|.
Now, if we draw a sketch of the situation, it is clear that both c and d are > 0 (c and d are above the X-axis and on the right side of the Y-axis). From the sketch it is also clear that d > c.
Therefore we can drop the absolute values:

|4d - 3c - 1| = |4c + 3d - 7|

4d - 3c - 1 = 4c + 3d - 7
and thus:

d = 7c - 6

Thus when we substitute d with 7c - 6:
$\displaystyle 4c +3(7c-6)-7 = 5 \sqrt{(2-c)^2 + (3-(7c-6))^2}$
$\displaystyle 4c +21c -18 -7 = 5 \sqrt{(2-c)^2 + (9-7c)^2}$
$\displaystyle 25c - 25 = 5 \sqrt{(2-c)^2 + (9-7c)^2}$
$\displaystyle (5c - 5)^2 = (2-c)^2 + (9-7c)^2$
$\displaystyle 25c^2 - 50c + 25 = 4 - 4c + c^2 +81 - 126c +49c^2$
$\displaystyle 25c^2 - 80c +60 = 0$

This equation has two solutions: c = 2 and c = 6/5
Substituting these values in d = 7c - 6 gives d = 8 and d =12/5

Thus the equations are:
$\displaystyle (x-2)^2 + (y-8)^2 = 25$
$\displaystyle (x-\frac{6}{5})^2 + (y- \frac{12}{5})^2 = 1$

6. You are right.

,

,

,

,

,

,

,

,

,

,

,

,

,

,

# what is the equation of a circle with two tangent line

Click on a term to search for related topics.