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The line y=x+1 intersects the circle at a and b.Find the exact length of a and b?

The centre for the circle is (1,-3) and the radius is 5. I looked online for help but can't seem to find anything.Can someone please tell me how to solve this or give me clues?
Reply 1
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number0
12
If 2 straight lines were intersecting what would you need to find the point of intersection (and you don't have a graph)?
Use Pythagoras
Reply 3
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LeonDH
9
Original post by Anonymous1502
The centre for the circle is (1,-3) and the radius is 5. I looked online for help but can't seem to find anything.Can someone please tell me how to solve this or give me clues?


Set up the simultaneous equations (x-1)^2 + (y+3)^2 = 25, and y = x=1.

Once you solve, you can use Pythagoras to find the length of ab
I'd start by drawing a sketch. I think the easiest thing here is to work out the perpendicular distance from the circle center to the line (*). If the foot of this perpendicular is P, then P bisects AB. And you can then use Pythagorus to deduce AP from the radius of the circle and the distance from P to the center (the value you found in (*)).
(edited 6 years ago)
Original post by number0
If 2 straight lines were intersecting what would you need to find the point of intersection (and you don't have a graph)?


The equation of one of the lines and then find what the perpendicular equation is, or you need at the least the coordinates of one of the points to find out the equation of one line if you don't have the equation already.Im not sure if this is right?
Reply 6
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number0
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I sorta meant you need the two equations of the lines, you don't know whether they are perpendicular or not. Solving these simultaneously would find the intersection. This can then be applied to this problem, you need the two equations to solve for their intersection(s!). you will have to create one of these with the given information.
Reply 7
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number0
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Once you have the coordinates of the 2 intersections then it's just down to finding the length
Original post by Anonymous1502
The centre for the circle is (1,-3) and the radius is 5. I looked online for help but can't seem to find anything.Can someone please tell me how to solve this or give me clues?

The equation of a circle of radius 5 centered on (1,-3) is:

(x−1)2+(y+3)2=52(x-1)^2+(y+3)^2=5^2

You have the line y=x+1y=x+1.

To find the intersection, substitute for yy (from the line's equation) in the circle's equation, and solve:

(x−1)2+(x+1+3)2=52(x-1)^2+(x+1+3)^2=5^2

Solving this equation will give you the xx values of the two intersections. You can get the yy values (most simply) from the equation of the line. Then use Pythagoras' theorem to get the length of AB.
Original post by LeonDH
Set up the simultaneous equations (x-1)^2 + (y+3)^2 = 25, and y = x=1.

Once you solve, you can use Pythagoras to find the length of ab


Thank you.I did what you said and I got x=1 and y=2. Would I then just you those to equal a and b so 1^2 + 2^2 =c^

1+4=5^2
so is the length just root 5?
Original post by RogerOxon
The equation of a circle of radius 5 centered on (1,-3) is:

(x−1)2+(y+3)2=52(x-1)^2+(y+3)^2=5^2

You have the line y=x+1y=x+1.

To find the intersection, substitute for yy (from the line's equation) in the circle's equation, and solve:

(x−1)2+(x+1+3)2=52(x-1)^2+(x+1+3)^2=5^2

Solving this equation will give you the xx values of the two intersections. You can get the yy values (most simply) from the equation of the line. Then use Pythagoras' theorem to get the length of AB.

So if I got the x value as 1 and -4 and y as 2 and -3 which values do i put in?Is it 1 and 2 because they're positive?
Original post by Anonymous1502
So if I got the x value as 1 and -4 and y as 2 and -3 which values do i put in?Is it 1 and 2 because they're positive?

x=1x=1 is the one point of intersection. Its yy value is y=x+1=2y=x+1=2. So, you have one point of intersection as (1,2)(1,2), which I'll call B.

The other point of intersection is (−4,−3)(-4,-3), which I'll call A (as its x value is lower).

Now work-out the change in x and y between the two points. With that, can you calculate the distance between them?
Reply 12
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LeonDH
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Original post by Anonymous1502
Thank you.I did what you said and I got x=1 and y=2. Would I then just you those to equal a and b so 1^2 + 2^2 =c^

1+4=5^2
so is the length just root 5?


No. There are two sets of solutions (-4, -3) and (1,2). a and b are the two points of intersection. I am assuming you are trying to find the the distance from point a to b. The change on the x-axis is 5, and the change on the y-axis is 5, so it is 5^2 + 5^2 = h^2.

h (the hypotenuse of the triangle) comes out as 5root2.

Does this make sense?
Original post by LeonDH
No. There are two sets of solutions (-4, -3) and (1,2). a and b are the two points of intersection. I am assuming you are trying to find the the distance from point a to b. The change on the x-axis is 5, and the change on the y-axis is 5, so it is 5^2 + 5^2 = h^2.

h (the hypotenuse of the triangle) comes out as 5root2.

Does this make sense?


That completely makes sense thanks :smile:
Original post by RogerOxon
x=1x=1 is the one point of intersection. Its yy value is y=x+1=2y=x+1=2. So, you have one point of intersection as (1,2)(1,2), which I'll call B.

The other point of intersection is (−4,−3)(-4,-3), which I'll call A (as its x value is lower).

Now work-out the change in x and y between the two points. With that, can you calculate the distance between them?


Thank you

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