Intersection of two circles in two points

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Rob Beckett

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Rob Beckett 6 years, 9 months ago

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Newcastle University Mathematics and Statistics 9 years, 2 months ago

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Name	Status	Author	Last Modified
Intersection of two circles in two points	Ready to use	Newcastle University Mathematics and Statistics	20/11/2019 14:50
Luis's copy of Intersection of two circles in two points	draft	Luis Hernandez	30/11/2018 15:37

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Question statement

Find the points where the circles:

$\simplify[std]{x^2+y^2 -{2*k2}x -{2*l2}y+{2*k2*a+2*l2*b-a^2-b^2}}=0$

and $\simplify[std]{x^2+y^2 -{2*k}x -{2*l}y+{2*k*a+2*l*b-a^2-b^2}}=0$

meet.

Enter the coordinates for the two points, starting with the point with the smaller $x$ -coordinate.

Note that, for example, $-3$ is smaller than $-1$ . Thus if the points of intersection were $(-1, 4)$ and $(-3,6)$ you would enter $(-3,6)$ for the first point and $(-1, 4)$ for the second.

Input the coordinates as integers.

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	-4
k2	rational	-6
c	integer	-8
b	integer	0
d	integer	-2
g	rational	2
f	rational	1/2
t2	integer	0
s1	integer	-1
m	integer	-8
l	rational	7
n	integer	-2
q	integer	0
p	integer	-4
l2	rational	-1
b1	integer	2
k	rational	-10
t	integer	4

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Generated value: `integer`

-4

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Advice

Subtract the equation for the first circle from the equation for the second. You get $\simplify[std]{{2*k2-2*k}x+ {2*l2-2*l}y+{2*k*a+2*l*b-2*k2*a-2*l2*b}}=0$ .

This can be rearranged as: $y = \simplify[std]{({b-d}/{a-c})x+{b*c-a*d}/{c-a}}$

Substitute for $y = \simplify[std]{({b-d}/{a-c})x+{b*c-a*d}/{c-a}}$ in the equation of the circle.

We get $\simplify[std]{x^2+(({b-d}/{a-c})x+{b*c-a*d}/{c-a})^2 -{2*k}x -{2*l}(({b-d}/{a-c})x+{b*c-a*d}/{c-a})+{2*k*a+2*l*b-a^2-b^2}}=0$ .

Collecting terms we get $\simplify[std]{{1+ (b-d)^2/(a-c)^2} x^2+{(-2)*(b-d)(b*c-a*d)/(a-c)^2-2*k -(2*l)*(b-d)/(a-c)}x+{(b*c-a*d)^2/(c-a)^2 -2*l*(b*c-a*d)/(c-a)+2*k*a+2*l*b-a^2-b^2}}=0$ .

Solving this quadratic equation for $x$ gives solutions $x=\var{a}$ and $x=\var{c}$ .

Substitute these values of $x$ into the equation $\simplify[std]{({b-d}/{a-c})x+{b*c-a*d}/{c-a}}$ to obtain the $y$ -coordinates: $y=\var{b}$ (when $x=\var{a}$ ) and $y=\var{d}$ (when $x=\var{c}$ ).

Enter the coordinate pairs in the order $(\var{m},\var{n})$ , $(\var{p}, \var{q})$ .

Give a worked solution to the whole question.

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Rob Beckett 6 years, 9 months ago

Newcastle University Mathematics and Statistics 9 years, 2 months ago

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Intersection of two circles in two points

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

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History

Comment

Checkpoint description

Rob Beckett 6 years, 9 months ago

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

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Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`