Solve Circles & Chords: Find Equations to Satisfy Conditions

[odolwa99]

With this question, I have worked out the correct answers (see the section bordered by BBB), but my original approach was to go by the 1st attempt (bordered by AAA). In the 1st attempt, the h/ k equation results in a single centre, rather than the 2 required to form the 2 separate circle equations. Can anyone help me spot what I need to do to get AAA to work, or is BBB the only approach that will work here?

Many thanks.

Homework Statement

Q. A circle intersects a line at the points a(-3, 0) & b(5, -4). The lines midpoint, m, is (1, -2). The distance from the centre of the circle to m is $\sqrt{5}$. Find the equations of the 2 circles that satisfy these conditions.

Homework Equations

The Attempt at a Solution

$|ab|=\sqrt{(-3-5)^2+(0+4)^2}=\sqrt{80}$
1/2 length of chord = $\sqrt{20}$
$|ac|^2=\sqrt{5}^2+\sqrt{20}^2=25\rightarrow |ac|=5$ radius

AAA
Slope of |ab|: $\frac{-4-0}{5+3}=\frac{-1}{2}$
Slope of perpendicular line M = 2
M equation: $y+2=2(x-1)\rightarrow 2x-y-4=0$
M contains the centre (h, k): 2h-k-4=0

For a 2nd equation containing h & k use (-3, 0):
$(h+3)^2+(k-0)^2=k^2\rightarrow h^2+6h+9=0\rightarrow (h+3)(h+3)=0\rightarrow h=-3$
For h = -3: $2h-k-4=0\rightarrow k=2(-3)-4\rightarrow k=-10$
Thus circle centre is (-3, -10)
AAA

BBB
If radius = 5, then centre of circle containing (-3, 0) = (2, 0)
Equation: $(x-2)^2+(y-0)^2=25\rightarrow x^2+y^2-4x-21=0$

If radius = 5, then centre of circle containing (5, -4) = (0, -4)
Equation: $(x-0)^2+(y+4)^2=25\rightarrow x^2+y^2+8y-9=0$
BBB

 

[barryj]

Where did this come from? (h+3)2+(k−0)2=k2

 

[odolwa99]

The example in the textbook makes use of the equation for a circle, (h-x)^2+(y-k)^2=r^2. They define the centre of the circle (h, k) and the radius = |k| = |ac|.
So, without (h, k), one of the 2 points of contact with the circle (I went with (-3, 0)) is subbed into the equation in order to try and determine what (h, k) is, using this and the earlier 2k -y -4 = 0 equation.

Is that what you were looking for? It's a bit tricky to lay things out clearly, for a question like this.

 

[SammyS]

With this question, I have worked out the correct answers (see the section bordered by BBB), but my original approach was to go by the 1st attempt (bordered by AAA). In the 1st attempt, the h/ k equation results in a single centre, rather than the 2 required to form the 2 separate circle equations. Can anyone help me spot what I need to do to get AAA to work, or is BBB the only approach that will work here?

Many thanks.

Homework Statement

Q. A circle intersects a line at the points a(-3, 0) & b(5, -4). The lines midpoint, m, is (1, -2). The distance from the centre of the circle to m is $\sqrt{5}$. Find the equations of the 2 circles that satisfy these conditions.

Homework Equations

The Attempt at a Solution

$|ab|=\sqrt{(-3-5)^2+(0+4)^2}=\sqrt{80}$
1/2 length of chord = $\sqrt{20}$
$|ac|^2=\sqrt{5}^2+\sqrt{20}^2=25\rightarrow |ac|=5$ radius

AAA
Slope of |ab|: $\frac{-4-0}{5+3}=\frac{-1}{2}$
Slope of perpendicular line M = 2
M equation: $y+2=2(x-1)\rightarrow 2x-y-4=0$
M contains the centre (h, k): 2h-k-4=0

For a 2nd equation containing h & k use (-3, 0):
$(h+3)^2+(k-0)^2=k^2\rightarrow h^2+6h+9=0\rightarrow (h+3)(h+3)=0\rightarrow h=-3$
For h = -3: $2h-k-4=0\rightarrow k=2(-3)-4\rightarrow k=-10$
Thus circle centre is (-3, -10)
AAA

BBB
If radius = 5, then centre of circle containing (-3, 0) = (2, 0)
Equation: $(x-2)^2+(y-0)^2=25\rightarrow x^2+y^2-4x-21=0$

If radius = 5, then centre of circle containing (5, -4) = (0, -4)
Equation: $(x-0)^2+(y+4)^2=25\rightarrow x^2+y^2+8y-9=0$
BBB

AAA is not correct.

In BBB:
Yes you have the correct equations for the circles, but you don't show how you arrived at them.

 

[barryj]

I don't get it. Where did this come from "the radius = |k| = |ac|". You have the equation of the circle correct but I don't see how radius is abs(k). Your BB to BBB makes sense but I don't get the part AAA to AAA.

 

[odolwa99]

I don't get it. Where did this come from "the radius = |k| = |ac|". You have the equation of the circle correct but I don't see how radius is abs(k). Your BB to BBB makes sense but I don't get the part AAA to AAA.

From the books example, |k| was shown as a separate line drawn to the x-axis, but equal to the radius length of |ac|. That particular example makes reference to the circle touching the x-axis, and I think the purpose of |k| was to connect a radial line to the axis and work it into the line equation. It's the only example I have to work from at the moment, so I most likely have taken this out of it's original context. I'll include the diagram I was referring to in an attachment.

AAA is not correct.

In BBB:
Yes you have the correct equations for the circles, but you don't show how you arrived at them.

I realize AAA is not arrived at correctly. It was my best guess/ attempt at the solution, based on the examples in the textbook. Can you please indicate what I need to do differently in order to arrive at BBB, algebraically. Thank you.

 

[ehild]

$|ab|=\sqrt{(-3-5)^2+(0+4)^2}=\sqrt{80}$
1/2 length of chord = $\sqrt{20}$
$|ac|^2=\sqrt{5}^2+\sqrt{20}^2=25\rightarrow |ac|=5$ radius

AAA
Slope of |ab|: $\frac{-4-0}{5+3}=\frac{-1}{2}$
Slope of perpendicular line M = 2
M equation: $y+2=2(x-1)\rightarrow 2x-y-4=0$
M contains the centre (h, k): 2h-k-4=0

For a 2nd equation containing h & k use (-3, 0):
[itex](h+3)^2+(k-0)^2=k^2

You wrote up the square of distance from a to the centre of the circle. It is equal to the radius squared (25). Instead of k^2 you should write r^2=25 in the last equation.

ehild