find an equation of a circle tangent to the lines 3x-4y+17=0, 4x+3y+6=0, and 3x+4y-41=0.?

Answer 1

3x - 4y + 17 = 0

3x - 4y + 17 - 3x = 0 - 3x

- 4y + 17 = -3x

- 4y + 17 - 17 = - 3x - 17

-4y = - 3x - 17

-4y/-4 = - 3x/- 4 - 17/- 4

y - 3x/4 + 17/4

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

4x + 3y + 6 = 0

4x + 3y + 6 - 4x = 0 - 4x

3y + 6 = - 4x

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

3y = - 4x - 6

3y/3 = - 4x/3 - 6/2

y = - 4x/3 - 2

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

3x + 4y - 41 = 0

3x + 4y - 41 - 3x = 0 - 3x

4y - 41 = - 3x

4y - 41 + 41 = - 3x + 41

4y = -3x + 41

4y/4 = - 3x/4 + 41/4

y = -3x/4 + 41/4

- - - - - - - -s-

Answer 2

Solve for each y:
y = (3x + 17)/4
y = -2(2x + 3)/3
y = -(3x - 41)/4

Thus, the circle passes through the points
(x, (3x + 17)/4)
(x, -2(2x + 3)/3)
(x, -(3x - 41)/4)

let (h,k) be the center of the circle, and r be the radius. Then the 3 points satisfy the equation
(x - h)² + (y - k)² = r²

Thus,
(x - h)² + ((3x + 17)/4 - k)² = r²
(x - h)² + (-2(2x + 3)/3 - k)² = r²
(x - h)² + (-(3x - 41)/4 - k)² = r²

Now, we have 4 variables: x, h, k and r. So we need one more equation to solve for all of them. Since we know that the radius is perpendicular to the tangent line, then
(slope of radius) = -1/(slope of any tangent line)

Let's take the first tangent line. The slope is 3/4. The slope of the radius is -4/3. Since the radius intersects the tangent line and passes through the center (h,k), then
y - k = -4/3(x - h) is an equation of the radius. Since, from the first tangent line y = (3x + 17)/4, then
(3x + 17)/4 - k = -4/3(x - h)
________________________________
Thus, we have 4 equations:
(x - h)² + ((3x + 17)/4 - k)² = r²
(x - h)² + (-2(2x + 3)/3 - k)² = r²
(x - h)² + (-(3x - 41)/4 - k)² = r²
(3x + 17)/4 - k = -4/3(x - h)

From these equations, we can solve for the values of h, k and r, and substitute the values to the original circle equation
(x - h)² + (y - k)² = r²

Then, you get the equation of your circle. Hope this helps your solution.

^_^

Answer 3

aileen_, this may be "cheating", but I notice the first two lines are perpendicular to each other, and so its centre must lie on the bisector of the angle between them. Actually, that may not help. Why not just use the sort of approach you took on the previous problem I responded to: Use
distance of centre from line = radius of circle,
and so if the circle is (x-h)^2 + (y-k)^2 = r^2, then

r^2 = ((3h-4k+17)^2)/(3^2+4^2), which becomes
25r^2 = 9h^2+16k^2+102h-136k+289. This looks a bit daunting, but if you do the same for the other two lines, they'll all have 25r^2 so you can eliminate that, and the third one will have the same coefficients of h^2 and k^2 as this one has so they'll drop out when you subtract. You'll have a linear equation from which to express k in terms of h and sub in the equation you get from 1 and 2. Again I'm going to post this now, and edit it a bit later when I've worked out the answer.

Later: Yeah, well that was horrible. There must be a neater way to do it. I found one centre is (-9, 14) with radius ?? depends on what line you take it from so obviously I made a mistake in all that messy algebra with big coefficients, and I haven't time to do it again. How about my other idea? The bisectors of the rightangles between the first two lines are
7x - y + 23 = 0 and x + 7y - 11 = 0

Unfortunately, looking at a sketch I think there are four solutions to this problem!! Let's just take the first bisector, so the centre satisfies
7h - k + 23 = 0.

Its distance from the third line is
abs((3h+4k-41)/5), and we want that to be equal to
abs((4h+3k+6)/5)

If you ignore abs you get h-k+47=0, and sub into this
k=7h+23, and get
h=4, k=51.

On the other hand if you sub h=11-7k into it, I think you get
h=-159/4, k= 29/4

If you now consider abs, allowing that
3h+k-41 = -4h-3k-6, you get
7h + 4k - 35 = 0. Sub k = 7h+23 and get
35h + 57 = 0 and so
h = -57/35, k = 58/35 [I think this is the centre of the circle which is inside the triangle formed by the three lines. The other solutions are all outside this triangle]

Instead sub h=11-7k and get
42 - 45k =0 and so
h=67/15, k=14/15

I haven't substituted any these centres into the equations of the three lines to see if the centre I've found really is equidistant from the lines. Maybe you can do that with one of them!

PS. Maybe you can see that Gopal's solution is wrong because he's found three different circles, but you need just one which touches all three lines. Kevin! has outlined a clever method, but it may well get into just as big a mess as I did when I tried the first time.Just check one of my answers to see if it's right!
Notice that although I said I didn't have time to do it again, I couldn't resist.

Answer 4

a) 3x - 4y + 17 = 0
-4y/4 = -3x/4 - 17/4
y = 3/4x + 17/4

b) 4x +3y + 6 = 0
3y/3 = -4x/3 - 6/3
y = -4/3x - 2

c) 3x + 4y - 41 = 0
4y/4 = -3x/4 + 41/4
y = -3/x4 + 41/4

Answer 5

the condition for y=mx+c to be a tangentto the circle x^2+y^2=a^2 is c^2=a^2(1+m^2)
equation 1
289=a^2(1+9/16)
a^2=289*16/25
so the equation is x^2+y^2=(68/5)^2
25x^2+25y^2=4624 or 68^2
equation2

36=a^2(1+16/9)
a^2=36*9/25
the equation is x^2+y^2=(18/5)^2
x^2+y^2=(18/5)^2
25x^2+25y^2=18^2
equation 3
1681=a^2(1+16/9)
a^2=1681*9/16
a^2=(123/4)^2
x^2+y^2=(123/4)^2
16x^2+16y^2=123^2

Answer 6

Why cant you find it?