# Thread: Really tough circle problem

1. ## Really tough circle problem

Hi, My problem is this : find the equation of the circle determined by the given conditions :

center (-2,3) and tangent to the line 4y-3x+2=0.

the equation for the line is y=3/4x -1/2 (im pretty sure)

so I drew that line, then I figured I would need to use the distance formula from where the line starting at (-2,3) with slope -4/3 (negative reciprical of 3/4) intersects 4y-3x+2=0.

However, it doesnt intersect at a point, its an arbitrary point. I cant determine the exact x,y of it by graphing. I dont see how to make the distance formula = to an arbitrary point. Also the only other thing I can think of is that the circle can only intersect 4y-3x+2=0 once not twice. I also cant figure out how to make any triangles that would help me. Im stumped. What Do I do? Who knows the answer? I want to know badly.

2. You almost have it. *Really* close.

Formula for line from (-2, 3) with slope -4/3 is
y - 3 = -4/3(x + 2)

Now find where this intersects
4y - 3x + 2 = 0

These should intersect at one point, P, because they are perpendicular.

Calculate the distance from (-2, 3) to P, and this is the radius for your circle.

This touches the line once. Why? The radius from (-2,3) to P forms right angles with the line.

3. The distance from $(-2,3)$ to the line $3x-4y-2=0$ is $\dfrac{|3(-2)-4(3)-2|}{\sqrt{(3)^2+(-4)^2}}$.
That distance is the radius of the circle.

4. I want to know how to find p. When I graphed it P was not a point I could see, it was between different coordinates. How can I get the distance between the center and P if I have to guess what P is because its not on an exact point (meaning x and y are fractions and I cant tell what the fractions are.)

Also I got another similar question. - Center on the line x+y=1, Passes through (-2,1) and (-4,3):

I think that -2,1 and -4,3 should be focus and directx of a parabola, and any point on that parabola that hits the line x+y=1 should be the center since it will be equal distance from the 2 points and also on the line.

I got (y-2)=1/4(x+3) as the equation of the parabola (used midpoint formula to get vertex) and y=-x+1 as the line.

but I have the same problem as the other question. The line intersects the parabola at a point I can not tell when I graph them. x and y are fractions not exact points. How do I figure out problems that bottle down like this?

I need to know how to find P basicly. I mean, its possible my math was just wrong and P turned out to be a point on the graph though that is actual numbers not fractions.

(also how do I thank people who try to help me? I see it has a thanked option)

5. When two lines intersect, their x and y coordinates are equal.

y - 3 = -4/3(x + 2) intersects 4y - 3x + 2 = 0

at the x and y that simultaneously solves the 2 equations.
This solution for x and y is P.

Also you can use the formula for distance from point to line formula provided in an earlier post (without directly finding P).

6. To answer the OP, all one needs is the distance from the center to the tangent line.
We do not need to find p.

7. Your parabola question is a bit unclear.
Did you mean (y-2) = 1/4(x+3)^2 ?

To find the intersection between the parabola and the line y=-x+1

Again, you find the simultaneous solution(s) for x and y that solve both
(y-2) = 1/4(x+3)^2 and y=-x+1

Example: You can use substitution to get
(y-2) = (-x+1-2) = -x-1 = 1/4(x+3)^2
solve for x
and then use this to find y = -x+1

8. yeah i meant ^2 on the end of that parabola.

Also, I realize that the answer is the distance between p and center.

By graphing the line 4y-3x+2=0 I can not clearly tell which point intersects with the perpendicular line from the center of the circle. Therefore how can I tell what to input in the distance formula? This is what I want to know.

OK to best clear this up in my mind, You can tell me what to fill into this equation :

(x - -2)^2 + (y - 3)^2 = Z

What is x and y and how did you get x and y.

(-2 and 3 are the center of the circle, x and y are the points I cant clearly see when i graph the equation 4y-3x+2=0. also I understand what point it should be (the intersection) I just cant tell what it is)

9. Alright, lets do it by the intersection method.

y - 3 = -4/3(x + 2) intersects 4y - 3x + 2 = 0 where?
Lets solve this system (substitue y = -4/3(x+2) + 3 into the second equation)
4(-4/3(x+2) + 3) - 3x + 2 = 0
Solving give x = 2/5
y = -4/3(2/5 + 2) + 3 = -1/5

so P = (2/5, -1/5)

Distance from (-2,3) to (2/5,-1/5)?
sqrt((-2 - 2/5)^2 + (3 + 1/5)^2) = 4

Equation for the circle at center -2, 3 with radius 4 is
(x + 2)^2 + (y - 3)^2 = 4^2

10. Originally Posted by snowtea
Your parabola question is a bit unclear.
Did you mean (y-2) = 1/4(x+3)^2 ?

To find the intersection between the parabola and the line y=-x+1

Again, you find the simultaneous solution(s) for x and y that solve both
(y-2) = 1/4(x+3)^2 and y=-x+1

Example: You can use substitution to get
(y-2) = (-x+1-2) = -x-1 = 1/4(x+3)^2
solve for x
and then use this to find y = -x+1
hold on sorry didnt see this post trying it now. Also plato I saw the formula you posted but I didnt get why it would work. never seen that before.

11. On a tangent line to a circle, the point of tangency is the nearest point to the center.
So the distance from the center to a tangent line is the radius of the circle.

12. Awsome ! I never thought to substitute. Oops. I guess that when you substitute it takes the 1st equation and just adds the parameters of the 2nd. Then when you solve bang. Cool I so get it now. thank you so much to both of you for helping!

Ps how do I thank you guys? I see you get creds for helping people.

13. Originally Posted by Plato
The distance from $(-2,3)$ to the line $3x-4y-2=0$ is $\dfrac{|3(-2)-4(3)-2|}{\sqrt{(3)^2+(-4)^2}}$.
That distance is the radius of the circle.
Why does dividing the absolute value of the center into the equation of the line (without the =0) by the square root of the constants to the line squared give you the distance of the radius of the circle? How you help me picture why that works ?

14. Glad you got it. I think you were already really close from the beginning.

You can thank people by clicking the thanks button (thumbs up icon) under the post.

15. The formula is a shortcut for what you did with the perpendicular line.
If you solve for P algebraically and calculate this distance, it simplifies to the formula.
You can derive the formula in this way.

The reason why the equation looks so simple requires understanding a bit about vectors and dot products. You don't have to worry about it if you haven't learned about vectors yet.

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