How to find the radii of these 2 circles given 2 known points

In summary, the homework statement is that two circles have their centers at (1.5, 3.5) and (2, 4). The first circle has a radius of -1 and the second circle has a radius of 1.5.
  • #1
Helly123
581
20

Homework Statement

Screenshot_34.png


Homework Equations



y-y1 = m (x-x1) ---> line equation

$$ (x - a)^2 + (y-b)^2 = r^2 $$ ---> circle equation

The Attempt at a Solution



I tried to draw the triangles using, (1, 3) (2, 4) and (0, b)

(0, b) is the tangent point to y-axisand used those points for making perpendicular bisector, so I can find the center of the circle, and find the distance from that center to one of the points, which that distance is the radii.between (1, 3) (2, 4)

the mid point = 1.5 , 3.5 gradient = 1, the perpendicular m = -1

the line equation = y - 1.5 = -1(x - 3.5) ... (1)between (2,4) (0, b)

$$ mid point = (1, \frac {4+b)} {2} )

\\gradient = \frac {4-b}{2} \\the perpendicular gradient = \frac {-2}{4-b}

\\the line equation = y - \frac {(4+b)} {2} = \frac {-2}{(4-b)} (x - 1) ...(2) $$between (1,3) (0,b)

$$ \\mid point = (0.5, \frac {(3+b)} {2} )

\\gradient = \frac {(3-b)}{1} \\the perpendicular gradient = \frac {-1}{(3-b)}

\\the line equation = y - \frac {(3+b)} {2} = \frac {-1}{(3-b)} (x - 0.5) ...(3)

$$if I find the x, y then its the center

we know that the center (x, b) where b the same in b (0, b)

but i don't think its work.
can anyone give me another way to solve?

I find another solutions but I don't get what the person doing can anyone explain it?
https://www.algebra.com/algebra/homework/Circles/Circles.faq.question.1077722.html
 
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  • #2
Because we are asked not to graph this, I would rule out most approaches based on geometry and go for something algebraic.

By inspection, the centers of both circles are at y=2. We do not know their respective x coordinates. So write down some equations.

Let ##(x_1,y_1)## be the center of the first circle and let ##r_1## be its radius.

The radius of the first circle is given by the distance from its center to the tangent point on the y axis:
$$r_1 = x_1$$
The radius of the first circle is also given by the distance from its center to point (1,3):
$$r_1^2 = (1-x_1)^2 + (3-y_1)^2$$
We already know that ##y_1 = 2##. We already have that ##r_1 = x_1##. Making those substitutions:

$$x_1^2 = (1-x_1)^2 + 1^2$$

Solve for ##x_1##

Repeat for the second circle.
 
Last edited:
  • #3
jbriggs444 said:
Because we are asked not to graph this, I would rule out most approaches based on geometry and go for something algebraic.

By inspection, the centers of both circles are at y=2. We do not know their respective x coordinates. So write down some equations.

Let ##(x_1,y_1)## be the center of the first circle and let ##r_1## be its radius.

The radius of the first circle is given by the distance from its center to the tangent point on the y axis:
$$r_1 = x_1$$
The radius of the first circle is also given by the distance from its center to point (1,3):
$$r_1^2 = (1-x_1)^2 + (3-y_1)^2$$
We already know that ##y_1 = 2##. We already have that ##r_1 = x_1##. Making those substitutions:

$$x_1^2 = (1-x_1)^2 + 1^2$$

Solve for ##x_1##

Repeat for the second circle.
how do you know the center y = 2? inspection of what?
 
  • #4
Helly123 said:
how do you know the center y = 2? inspection of what?
My mistake. I completely misread the problem somehow.
 
  • #5
Helly123 said:
but i don't think its work.
You seemed to be doing ok. It should get you there.

What do you not understand in the solution you linked?
 
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  • #6
haruspex said:
What do you not understand in the solution you linked?
From the first, finding center of upper circle , how it comes up (1.5 -P/root 2 , 3.5 + P/root 2) ?
And how to find radii? R^2 = (1.5 -P/root 2)^2 + (3.5 + P/root 2)^2 not the same answer as from the link

While my method i get imaginary x
 
  • #7
Helly123 said:
From the first, finding center of upper circle , how it comes up (1.5 -P/root 2 , 3.5 + P/root 2) ?
And how to find radii? R^2 = (1.5 -P/root 2)^2 + (3.5 + P/root 2)^2 not the same answer as from the link

While my method i get imaginary x
What is P?

I would have done this:
Centre of circle is at (R,b). This is distance R from each of the two given points...
 
  • #8
haruspex said:
What is P?

I would have done this:
Centre of circle is at (R,b). This is distance R from each of the two given points...

So i used your method
$$ (R-1)^2 + (b-3)^2 = (R-2)^2 + (b-4)^2 $$
R + b = 5
b = 0 or 4
R = 5 or 1
So does it mean the center of circle A = R, b = 1, 4
And circle B = R, b = 5, 0
??
 

Related to How to find the radii of these 2 circles given 2 known points

1. How do you find the radii of two circles given two known points?

The radii of two circles can be found by using the distance formula. First, calculate the distance between the two known points using the formula: √[(x2 - x1)^2 + (y2 - y1)^2]. This will give you the distance between the centers of the two circles. Then, use the distance to one of the known points as the radius for each circle.

2. Can the radii of two circles be different if the two known points are the same?

Yes, the radii of two circles can be different even if the two known points are the same. This is because the two circles may have different centers and therefore, different distances from the known point.

3. What if one of the known points is the center of one of the circles?

If one of the known points is the center of one of the circles, then the radius for that circle will be 0. This is because the distance from the center to itself is 0.

4. Can you find the radii of two circles if only one known point is given?

No, the radii of two circles cannot be found if only one known point is given. This is because you need at least two points to determine the distance between the centers of the two circles.

5. Are there any other methods for finding the radii of two circles given two known points?

Yes, another method is to use the equation of a circle: (x - h)^2 + (y - k)^2 = r^2. Plug in the coordinates of the two known points and solve for r. However, this method may be more complex and may not work for all cases.

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