How do you work out where this circle intercepts this line?

Question

A Circle, radius 3 centered at the point (1, 1) and the line y=2x+1

I already know the formula r^2 = (X-a)^2 + (mx+c-b)^2

I just can't understand how to getan answer. (Please answer in steps so i can understand easier).

Answer 1

The circle equation:
(x-1)²+(y-1)²=9 .................. (I)

The line equation:
f(x) = y=2x+1 ..................... (II)

You have to put (II) in (I):
(x-1)²+(2x+1-1)²=9
(x-1)²+(2x)²=9
x²-2x+1+4x²=9
5x²-2x-8 = 0

x = [2±(2²+4*5*8)^½] / (2*5)
x = [2±(4+160)^½] / 10
x = (2±2*41^½) / 10

x = (1±41^½) / 5
=>
x' = (1-41^½) / 5 ......... x'' = (1+41^½) / 5

The intersection points are:
M((1-41^½)/5, f((1-41^½)/5))

and

N((1+41^½)/5, f((1+41^½)/5))

₢

Answer 2

just remember that your two formulas that you used
y = mx +c for the line
r^2 = (x-a)^2 + (y-b)^2 for the circle

are generic, or for any circle and line.
you already have m and c and r and a and b
your line and circle are

y = 2x+1
3^2 = (x-1)^2 + (y-1)^2

Now the formula you used was just replacing the 'y' in the circle by '2x+1' because y=2x+1 from the line.

So you need to solve:

3^2 = (x-1)^2 + (2x+1-1)^2
you will notice that there are only x's in there. so work out what x is.

9 = x^2 - 2x +1 + 4x^2
0 = 5x^2 - 2x -8
0 = (5x-? )(x+? )... hmmm not working...
ok the hard way:

x = (-2ab +- strq())/(
hmmmm, I don't remember the formula...
anyway, work out the X's (you should get 2 the line will cross the circle at 2 points)
and the work out what the Y is (using y = 2x+1)

and then you have your points.

sorry for the incomplete answer

Answer 3

Centre is (1,1) and radius is 3 ....
so equation of the circle is (x-1)*2 + (y-1)*2 = 9
The line represented by y = 2x + 1
To find the interception point, just solve these two equations. i.e. substitute y = 2x +1 in the circle equation. That gives you .....
(x-1)*2 + 4x*2 = 9
x*2 - 2x +1 + 4x*2 = 9
5x*2 - 2x -8 = 0.
Solve this equation for x and from that find y. That gives you the point of intersection.

Answer 4

Circle formula
(x - 1)^2 + (y - 1)^2 = 3^2
x^2 - 2x + 1 + y^2 - 2y + 1 = 9
x^2 + y^2 - 2x - 2y -7 = 0 (This is your circle equation)

(Substituting in for 'Y')
x^2 + (2x + 1)^2 - 2x - 2(2x + 1) - 7 = 0
x^2 + 4x^2 + 4x + 1 - 2x -4x - 2 -7 = 0 (Collect like terms)
5x^2 - 2x - 8 = 0
Using x = - b +/- /(b^2 - 4ac)
_______________
2a

X = - - 2 +- /((-2)^2 - 4(5)(-8))
_____________________
2(5)
X = 2 +/- /(4 + 160)
_____________
10

X = 2 +/- /164
________

10

x = 2 +/- 12.806
__________
10

x = 2 + 12.806 & x = 2 - 12.806
_________ ________
10 10
x = 1.4806 & -1.0806

Using the straight line equation., and substituting in for 'x'.

y = 2(1.4806) + 1 & y = 2(-1.0806) + 1
y = 3.9612 y = -1.1612

Hence the Straight line intercepts the circle at two points :-

(1.4806, 3.9612) & (-1.0806, -1.1612)

Answer 5

Draw a graph numbered 1,2,3,4,5 and both the X and Y axis.

Draw a rough circle with centre point 1,1 and a radius of three. this means draw it so it touches x3,y3 on the graph.

Take values for x1,x2,y1,y2. within the range of the circle.

Calculate th slope using the Formula (y2-y1)/(x2-x1)

Now you have a value for x in the formula y=2x+1.

So solve for y. (i did this myself, the y value turns out to be 5)

Now you have a value for x(2) and y(5).

Mark these on your graph and draw a straight line through the points.

And now you can see where the circle intercepts the line.

Answer 6

hi anu you find out the point of intersection by solving both the equations simultaneously
equation of the circle is x^2-2x+y^2-2y=7
sub y=2x+1, the equation of the line and solving
x^2-2x+(2x+1)^2-2(2x+1)=7
x^2-2x+4x^2+4x+1-4x-2=7
5x^2-2x-8=0
x=[2+/-rt(4+160)]/10
=2+-12.8/10
=-1.08 or 1.48
corresponding valuesof y will be -1.16 and 3.96
so the points of intersection are
{-1.08,-1.16};{1.48,3.96}