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Just solve y=3x-1 and ((x-4)^2)+((y-1)^2)=9 as simultaneous equations:

y=3x-1

=> ((x-4)^2)+((3x-1-1)^2)=9

=> ((x-4)^2)+((3x-2)^2)=9

=>(x^2)-8(x)+16+9(x^2)-12x+4=9

=> 10(x^2)-20x+11=0 (i)

Now find the discriminant (D=(b^2)-4ac) of (i):

D=400-4(10)(11)

Therefore:

D=-40

Hopefully you recognize D as a part of the quadratic formula; since you cannot have a real square root of a negative number-this equation has no real roots, and hence the circle and the line do not intersect.

The second part would be very stupid to ask as it can be a tangent to an infinite number of circles with their centroids being placed at infinte possibilities along a certain line.

Newton.

y=3x-1

=> ((x-4)^2)+((3x-1-1)^2)=9

=> ((x-4)^2)+((3x-2)^2)=9

=>(x^2)-8(x)+16+9(x^2)-12x+4=9

=> 10(x^2)-20x+11=0 (i)

Now find the discriminant (D=(b^2)-4ac) of (i):

D=400-4(10)(11)

Therefore:

D=-40

Hopefully you recognize D as a part of the quadratic formula; since you cannot have a real square root of a negative number-this equation has no real roots, and hence the circle and the line do not intersect.

The second part would be very stupid to ask as it can be a tangent to an infinite number of circles with their centroids being placed at infinte possibilities along a certain line.

Newton.

Newton

Just solve y=3x-1 and ((x-4)^2)+((y-1)^2)=9 as simultaneous equations:

y=3x-1

=> ((x-4)^2)+((3x-1-1)^2)=9

=> ((x-4)^2)+((3x-2)^2)=9

=>(x^2)-8(x)+16+9(x^2)-12x+4=9

=> 10(x^2)-20x+11=0 (i)

Now find the discriminant (D=(b^2)-4ac) of (i):

D=400-4(10)(11)

Therefore:

D=-40

Hopefully you recognize D as a part of the quadratic formula; since you cannot have a real square root of a negative number-this equation has no real roots, and hence the circle and the line do not intersect.

The second part would be very stupid to ask as it can be a tangent to an infinite number of circles with their centroids being placed at infinte possibilities along a certain line.

Newton.

y=3x-1

=> ((x-4)^2)+((3x-1-1)^2)=9

=> ((x-4)^2)+((3x-2)^2)=9

=>(x^2)-8(x)+16+9(x^2)-12x+4=9

=> 10(x^2)-20x+11=0 (i)

Now find the discriminant (D=(b^2)-4ac) of (i):

D=400-4(10)(11)

Therefore:

D=-40

Hopefully you recognize D as a part of the quadratic formula; since you cannot have a real square root of a negative number-this equation has no real roots, and hence the circle and the line do not intersect.

The second part would be very stupid to ask as it can be a tangent to an infinite number of circles with their centroids being placed at infinte possibilities along a certain line.

Newton.

I UNDERSTAND THE ABOVE!cheers. well the second part is a question like this:

Prove that the line y=3x+1 is a tangent to the circle x^2 +y^2 -14x-4y+13=0. on the homework sheet..so how do i go about doing this. Oh and by the way the above, if i solve the answer using the discriminant, if i get a negative number does this mean theres no real roots and it isnt touching the circle, but if i got a postive that means it is? cheers for help.

Oh ok. Well look remember these rules for the discriminant:

((b^2)-4ac)=0 => One real root

((b^2)-4ac)>0 => Two real-distinct (seperate) roots

((b^2)-4ac)<0 => No real-distinct roots

((b^2)-4ac)=0 => One real root => Tthis root can also just be a tangent i. e. just touching at one point.

So let's look at your question. I'm really sorry but once again we have to use the ultimately boaring simultaneous equations:

y=3x+1 (i)

AND

(x^2)+(y^2) -14(x)-4(y)+13=0 (ii)

Substitute (i) into (ii)

(x^2)+((3x+1)^2) -14(x)-(4(3x+1))+13=0

(x^2)+(9(x^2))+6(x)+1-14(x)-12(x)-4+13=0

=> 10(x^2)-20(x)+10=0

Now find the discriminant, D:

D=((b^2)-4ac)=((20^2)-4(10)(10))=400-400=0

Now because we have only one root, which I have computed aside to be (1,4), this means that (3x-1) is a tangent to your circle.

Ok imagine a parabola, which has a turning point as soon as it touches the x-axis. That "touch-point" is a root. In that case D of that function would be equal to zero as well.

BASICALLY: ((b^2)-4ac)=0 => ONE ROOT => TOUCHING POINT => TANGENT.

Hope this helps!

Newton.

((b^2)-4ac)=0 => One real root

((b^2)-4ac)>0 => Two real-distinct (seperate) roots

((b^2)-4ac)<0 => No real-distinct roots

((b^2)-4ac)=0 => One real root => Tthis root can also just be a tangent i. e. just touching at one point.

So let's look at your question. I'm really sorry but once again we have to use the ultimately boaring simultaneous equations:

y=3x+1 (i)

AND

(x^2)+(y^2) -14(x)-4(y)+13=0 (ii)

Substitute (i) into (ii)

(x^2)+((3x+1)^2) -14(x)-(4(3x+1))+13=0

(x^2)+(9(x^2))+6(x)+1-14(x)-12(x)-4+13=0

=> 10(x^2)-20(x)+10=0

Now find the discriminant, D:

D=((b^2)-4ac)=((20^2)-4(10)(10))=400-400=0

Now because we have only one root, which I have computed aside to be (1,4), this means that (3x-1) is a tangent to your circle.

Ok imagine a parabola, which has a turning point as soon as it touches the x-axis. That "touch-point" is a root. In that case D of that function would be equal to zero as well.

BASICALLY: ((b^2)-4ac)=0 => ONE ROOT => TOUCHING POINT => TANGENT.

Hope this helps!

Newton.

Newton

Oh ok. Well look remember these rules for the discriminant:

((b^2)-4ac)=0 => One real root

((b^2)-4ac)>0 => Two real-distinct (seperate) roots

((b^2)-4ac)<0 => No real-distinct roots

((b^2)-4ac)=0 => One real root => Tthis root can also just be a tangent i. e. just touching at one point.

So let's look at your question. I'm really sorry but once again we have to use the ultimately boaring simultaneous equations:

y=3x+1 (i)

AND

(x^2)+(y^2) -14(x)-4(y)+13=0 (ii)

Substitute (i) into (ii)

(x^2)+((3x+1)^2) -14(x)-(4(3x+1))+13=0

(x^2)+(9(x^2))+6(x)+1-14(x)-12(x)-4+13=0

=> 10(x^2)-20(x)+10=0

Now find the discriminant, D:

D=((b^2)-4ac)=((20^2)-4(10)(10))=400-400=0

Now because we have only one root, which I have computed aside to be (1,4), this means that (3x-1) is a tangent to your circle.

Ok imagine a parabola, which has a turning point as soon as it touches the x-axis. That "touch-point" is a root. In that case D of that function would be equal to zero as well.

BASICALLY: ((b^2)-4ac)=0 => ONE ROOT => TOUCHING POINT => TANGENT.

Hope this helps!

Newton.

((b^2)-4ac)=0 => One real root

((b^2)-4ac)>0 => Two real-distinct (seperate) roots

((b^2)-4ac)<0 => No real-distinct roots

((b^2)-4ac)=0 => One real root => Tthis root can also just be a tangent i. e. just touching at one point.

So let's look at your question. I'm really sorry but once again we have to use the ultimately boaring simultaneous equations:

y=3x+1 (i)

AND

(x^2)+(y^2) -14(x)-4(y)+13=0 (ii)

Substitute (i) into (ii)

(x^2)+((3x+1)^2) -14(x)-(4(3x+1))+13=0

(x^2)+(9(x^2))+6(x)+1-14(x)-12(x)-4+13=0

=> 10(x^2)-20(x)+10=0

Now find the discriminant, D:

D=((b^2)-4ac)=((20^2)-4(10)(10))=400-400=0

Now because we have only one root, which I have computed aside to be (1,4), this means that (3x-1) is a tangent to your circle.

Ok imagine a parabola, which has a turning point as soon as it touches the x-axis. That "touch-point" is a root. In that case D of that function would be equal to zero as well.

BASICALLY: ((b^2)-4ac)=0 => ONE ROOT => TOUCHING POINT => TANGENT.

Hope this helps!

Newton.

wow you are such a great help thank you i feel so dumb but i understand it all!Thank you lots

gemma.....

wow you are such a great help thank you i feel so dumb but i understand it all!Thank you lots

Oh just another thing, what would i do with chords?

This is prove that part of the line:

y=3x+5 forms a chord to the circle x^2+y^2-2x-6y+5=0 , also ifnd the length of this chord.

So what would i do solve this all simultaneously//and then how would i find the length of it?

gemma.....

Oh just another thing, what would i do with chords?

This is prove that part of the line:

y=3x+5 forms a chord to the circle x^2+y^2-2x-6y+5=0 , also ifnd the length of this chord.

So what would i do solve this all simultaneously//and then how would i find the length of it?

This is prove that part of the line:

y=3x+5 forms a chord to the circle x^2+y^2-2x-6y+5=0 , also ifnd the length of this chord.

So what would i do solve this all simultaneously//and then how would i find the length of it?

x^2 + y^2 - 2x - 6y + 5 = 0

x^2 - 2x + (y-3)^2 - 9 + 5 = 0

x^2 - 2x + (3x+5-3)^2 - 9 + 5 = 0

x^2 - 2x + 9x^2 + 12x + 4 - 9 + 5 = 0

10x^2 - 10x = 0

x^2 + x = 0

b^2 - 4ac = 1^2 - 4*1*0 = 1 > 0

so there are two roots, so the line crosses twice and forms a chord

to find the length, find the coordinates A and B where the line crosses the circle:

x^2 + x = 0

x(x + 1) = 0

x = 0 or -1

y = 3x+5

y = 5 or 2

A(0,5) B(-1,2)

Length^2 = (-1-0)^2 + (2-5)^2

= 1 + 9

= 10

length = sqrt 10

I think.

gemma.....

y=3x+5 forms a chord to the circle x^2+y^2-2x-6y+5=0 , also ifnd the length of this chord.

So what would i do solve this all simultaneously//and then how would i find the length of it?

So what would i do solve this all simultaneously//and then how would i find the length of it?

Line: y = 3x + 5

Circle: x^2 + y^2 - 2x - 6y + 5 = 0

Substitute the equation for the line into the equation for the circle:

---> x^2 + (3x + 5)(3x + 5) - 2x - 6(3x + 5) + 5 = 0

---> x^2 + 9x^2 + 30x + 25 - 2x - 18x - 30 + 5 = 0

---> 10x^2 + 10x + 0 = 0

---> x^2 + x + 0 = 0

---> Discriminant = 1^2 - 4(1)(0) = 1 > 0

---> As the discriminant is > 0

---> There are 2 distinct real roots and hence 2 intersections between the line and curve and hence the line is a chord to the circle.

Now: x^2 + x = 0

---> x(x + 1) = 0

---> x = 0 OR x = -1

When x = 0: y = 0 + 5 = 5

---> Intersection at (0, 5)

When x = -1: y = 3(-1) + 5 = 2

---> Intersection at (-1, 2)

Length of Chord = Distance between 2 points of intersection

= Sqrt [(0--1)^2 + (5-2)^2]

= Sqrt [1 + 3^2]

= Sqrt 10

---> Length Of Chord = Sqrt 10

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