How do you find the intersection between $y=x-8$ and $x^2+y^2= 34$ ?

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How do you find the intersection between $y=x-8$ and $x^2+y^2= 34$ ?

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Answer 1 · 2016-11-25T21:32:58

Points of intersection are:

$(x_1,y_1) = (3 ,-5)$

$(x_2,y_2)=(5,-3)$

Explanation:

$x^2+y^2=34$ is the equation of a circle or radius $sqrt(34)$ and has its centre at the origin.

Expressing this with $y$ being the dependant variable we have:

$y=+-sqrt(34-x^2)$ ...................Equation(1)

We also have the straight line equation

$y=x-8$ ......Equation(2)

substitute for $y$ in equation(1) using equation(2) we have:

$x-8=+-sqrt(34-x^2)$

Squaring both sides

$(x-8)^2=34-x^2$

$x^2-16x+64=34-x^2$

$2x^2-16x+30=0$

Divide both sides by 2

$x^2-8x+15=0$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$x=(-(-8)+-sqrt( (-8)^2-4(1)(15)))/(2(1))$

$x=4+-1$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Using equation(2)

At $x=3 -> y=x-8 = -5$

At $x=5->y=x-8=-3$

Points of intersection are:

$(x_1,y_1) = (3 ,-5)$

$(x_2,y_2)=(5,-3)$

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How do you find the intersection between $y=x-8$ and $x^2+y^2= 34$ ?

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Explanation:

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How do you find the intersection between y=x-8y=x-8 and x^2+y^2= 34x^2+y^2= 34?

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How do you find the intersection between $y=x-8$ and $x^2+y^2= 34$ ?